Chapter # 35
Magnetic Field due to A current
SOLVED EXAMPLES
Example 35.1 A wire placed along north-south direction carries a current of 10 A from south to north. Find the magnetic field due to a 1 cm piece of wire at a point 200 cm north-east from the piece. Sol. The situation in shown in figure. As the distance of P from the wire is much larger than the length of the wire, we can treat the wire as a small element. The magnetic field is given by d l r 0 dB i 4 r 3 or,
dB
0 dl sin i 4 r2
(10 2 m) sin 45º 1 T m 10 = (10 A) A (2m)2 = 1.8 × 10–9 T. The direction of B is the same as that of d l r . From the figure, it is vertically downward. Example 35.2 Figure shows two long, straight wire carrying electric currents in opposite directions. The separation between the wires is 5.0 cm. Find the magnetic field at a point P midway between the wires. Sol.
The right hand thumb rule shows that the magnetic field at P due to each of the wires is perpendicular to the plane of the diagram and is going into it. The magnitude of the field due to each wires is
7 T m 2 10 (10 A ) A = 2.5 10 2 m
0i B= 2d
= 80 T. The net field due to both the wires is 2 × 80 T = 180 T. Exercise 35.3 Two long, straight wires, each carrying an electric current of 5.0 A, are kept parallel to each other at a separation of 2.5 cm. Find the magnitude of the magnetic force experienced by 10 cm of a wire. Sol.
The field at the site of one wire due to the other is 7 T m 2 10 (5 . 0 A ) A 0i B= = = 4.0 × 10–6 T. 2d 2.5 10 2 m
The force experienced by 10 cm of this wire due to the other is F = i B = (5.0 A) (10 × 10–2 m) (4.0 × 10–5 T) = 2.0 × 10–5 N. Example 35.4 A circular coil of radius 1.5 cm carries a current of 1.5 A. If the coil has 25 turns, find the magnetic field at the centre. Sol.
The magnetic field at the centre due to each turn is
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Chapter # 35
Magnetic Field due to A current 0i . 2
The net field due to all 25 times is
B=
0i n = 2
7 T m 2 10 (1.5 A ) 25 A 1.5 10 2 m
= 1.57 × 10–3 T. Example 35.5 A long solenoid is formed by winding 20 turns/cm. What current is necessary to produce a magnetic field of 20 mT inside the solenoid? Sol.
The magnetic field inside the solenoid is B = 0 ni or, or,
7 T m × (20 × 102 m–1 ) i 20 × 10–3 T = 4 10 A
i = 8.0 A.
QUESTIONS 1. 2. 3. 4.
5.
6. 7. 8. 9. 10.
11. 12.
FOR
SHORT
ANSWER
An electric current flows in a wire from north to south. What will be the direction of the magnetic field due to this wire at a point east of the wire ? West of the wire ? Vertically above the wire ? Vertically below the wire? The magnetic field due to a long straight wire has been derived in terms of 0, i and d. Express this in terms of 0 , i and d. You are facing a circular wire carrying an electric current. The current is clockwise as seen by you. Is the field at the centre coming towards you or going away from you ? In Ampere's law Bd 0i , the current outside the curve is not included on the right hand side. Does it
mean that the magnetic field B calculated by using Ampere's law, gives the contribution of only the currents crossing the area bounded by the curve ? The magnetic field inside a tightly wound, long solenoid is B = 0ni. It suggests that the field does not depend on the total length of the solenoid the field should not increase. Explain qualitatively why the extra–added loops do not have a considerable effect on the field inside the solenoid. A long, straight wire carries a current. Is Ampere's law valid for a loop that does not enclose the wire ? That encloses the wire but is not circular ? A straight wire carrying an electric current is placed along the axis of a uniformly charged ring. Will there be a magnetic force on the wire if the ring starts rotating about the wire ? If yes, in which direction ? Two wires carrying equal currents i each, are placed perpendicular to each other, just avoiding a contact. If one wire is held fixed and the other is free to move under magnetic forces, what kind of motion will result ? Two proton beams going in the same direction repel each other whereas two wires carrying currents in the same direction attract each other. Explain. In order to have a current in a long wire, it should be connected to a battery or some such device. Can we obtain the magnetic field due to a straight, long wire by using Ampere's law without mentioning this other part of the circuit ? Quite often, connecting wires carrying currents in opposite directions are twisted together in using electrical appliances. Explain how it unwanted magnetic field. Two current–carrying wires may attract each other. In absence of other forces, the wires will move towards each other increasing the kinetic energy. Does it contradict the fact that the magnetic force cannot do any work and hence cannot increase the kinetic energy ?
Objective - I
1.
A vertical wire carries a current in upward direction. An electron beam sent horizontally towards the wire will be deflected (A) towards right (B) towards left (C*) upwards (D) downwards manishkumarphysics.in
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Chapter # 35
Magnetic Field due to A current
,d m/okZ/kj rkj esa Åij dh vksj /kkjk izokfgr gks jgh gSA rkj dh vksj {kSfrt fn'kk esa Hkstk tk jgk bysDVªkuW iat q fo{ksfir gksxk (A) nka;h vksj (B) cka;h vksj (C*) Åij dh vksj (D) uhps dh vksj 2.
A current-carrying, straight wire is kept along the axis of a circular loop carrying a current. The straight wire (A) will exert an inward force on the circular loop (B) will exert an outward force on the circular loop (C*) will not exert any force on the circular loop (D) will exert a force on the circular loop parallel to itself. ,d /kkjkokgh oy; dh v{k ds vuqfn'k ,d lh/kk /kkjkokgh rkj j[kk gqvk gSA lh/kk rkj (A) o`Ùkkdkj oy; ij vUnj dh vksj cy yxk;sxkA (B) o`Ùkkdkj oy; ij ckgj dh vksj cy yxk;sxkA (C*) o`Ùkkdkj oy; ij dksbZ cy ugha yxk;sxkA (D) o`Ùkkdkj oy; ij vius lekukUrj cy yxk;sxkA
3.
A proton beam is going from north to south and an electron beam is going from south to north. Neglection the earth’s magnetic field, the electron beam will be deflected (A*) towards the proton beam (B) away from the proton beam (C) upwards (D) downwards
,d izkVs kWu iqt a mÙkj ls nf{k.k dh vksj xfr'khy gS rFkk ,d bysDVªkuW iat q nf{k.k ls mÙkj dh vksj xfr'khy gSA i`Foh ds pqEcdh; {ks=k dks ux.; ekuus ij] bysDVªkWu iaqt fo{ksfir gksxk (A*) izksVkWu iaqt dh vksj (B) izksVkWu iqat ls ijs (C) Åij dh vksj (D) uhps dh vksj 4.
A circular loop is kept in that vertical plane which contains the north-south direction. It carries a current that is towards north at the topmost point. Let A be a point on axis of the circle to the east of it and B a point on this axis to the west of it. The magnetic field due to the loop (A) is towards east at A and towards west at B (B) is towards west at A and towards east at B (C) is towards east at both A and B (D*) is towards west at both A and B
,d o`Ùkkdkj ywi mÙkj&nf{k.k fn'kk esa fLFkr m/okZ/kj ry esa j[kk gqvk gSA blds mPpre fcUnq ij /kkjk dh fn'kk mÙkj dh vksj gSA ekuk fd bldh v{k ij fcUnq A iwoZ dh vksj rFkk fcUnq B if'pe dh vksj gSA ywi ds dkj.k pqEcdh; {ks=k (A) A ij iwoZ dh vksj rFkk B ij if'pe dh vksj gSA (B) A ij if'pe vksj rFkk B ij if'pe dh vksj gSA (C) A rFkk B nksuksa ij iwoZ dh vksj gSA (D*) A rFkk B nksuksa ij if'pe dh vksj gSA 5.
Consider the situation shown in fig. The straight wire is fixed but the loop can move under magnetic force. The loop will -
fp=k esa iznf'kZr fLFkfr ij fopkj dhft;sA lh/kk rkj dlk gqvk gS] fdUrq ywi pqEcdh; cy ds izHkko esa xfr dj ldrk gSA ywi -
(A) remain stationery (C) move away from the wire (A) fLFkj jgsxk (C) rkj ls ijs xfr djsxk
(B*) move towards the wire (D) rotate about the wire (B*) rkj dh vksj xfr djsxk (D) rkj ds ifjr% ?kw.kZu djsxkA
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Chapter # 35 Magnetic Field due to A current 6. A charge particle is moved along a magnetic field line. The magnetic force on the particle is (A) along its velocity (B) magnetic field only (C) both of them (D*) none of them
,d vkosf'kr d.k dks pqEcdh; cy js[kk ds vuqfn'k xfr djok;h tkrh gSA d.k ij pqEcdh; cy gksxk (A) blds osx ds vuqfn'k (B) dsoy pqEcdh; {ks=k (C) ;g nksuksa gh (D*) buesa ls dksbZ ugha 7.
8.
A moving charge produces (A) electric field only (B) magnetic filed only ,d xfr'khy vkos'k mRiUu djrk gS (A) dsoy fo|qr {ks=k (B) dsoy pqEcdh; {ks=k
(C*) both of them
(D) none of these
(C*) ;g
(D) buesa
nksuksa gh
ls dksbZ ugha
A particle is projected in a plane perpendicular to a uniform magnetic field. The area bounded by a the path described by the particle is proportional to (A) the velocity (B) the momentum (C*) the kinetic energy (D) none of these
,d le:i pqEcdh; {ks=k ds ry ds yEcor~ ,d d.k iz{ksfir fd;k tkrk gSA d.k ds }kjk r; fd;s x;s iFk dk {ks=kQy lekuqikrh gksxk (A) osx ds (B) laosx ds (C*) xfrt ÅtkZ ds (D) buesa ls dksbZ ugha 9.
Two particles X and Y having equla charge, after being acceleration through the same potential difference circular paths of radii R1 and R2 respectively. The ratio of the mass of X to that of Y is ,d leku vkos'k okys nks d.k x rFkk y ,d leku foHko ls Rofjr gksus ds i'pkr~ le:i pqEcdh; {ks=k esa izfo"V gksrs gSa rFkk Øe'k% R1 ,oa R2 f=kT;k ds o`Ùkkdkj iFkksa ij xfr djrs gSAa x rFkk y ds nzO;ekuksa dk vuqikr gS (A) (R1/R2)1/2 (B) R1/R2 (C*) (R1/R2)2 (D) R1R2
10.
Two parallel wires carry currents of 20 A and 40 A in opposite directions. Another wire carrying a current antiparallel to 20 A is placed midway between the two wires. The magnetic force on it will be (A) towards 20 A (B*) towares 40 A (C) zero (D) perpendicular to the plane of the current nks lekukUrj rkjksa esa 20 A rFkk 40 A /kkjk,¡ ijLij foijhr fn'kkvksa esa izokfgr gks jgh gSA bu nksuksa rkjksa ds Bhd e/; esa ,d vU; rkj fLFkr gSA ftlls /kkjk dh fn'kk 20 A okys rkj dh /kkjk ls foifjr gSA bl ij yxus okyk pqEcdh; cy gksxk (A) 20 A dh vksj (B*) 40 A dh vksj (C) 'kwU; (D) /kkjkvksa ds ry ds yEcor~
11.
Two parallel, long wires carry currents i1 and i2 with i1 > i2. When the current are in the same direction, the magnetic field at a point midway between the wire is 10T. If the direction of i2 is reversed, the field becomes
i1 30T. The ratio i is 2
nks lekUrj ,oa yEcs rkjksa esa i1 rFkk i2 /kkjk,¡ izokfgr gks jgh gS rFkk i1 > i2 gSA tc /kkjk,¡ ,d gh fn'kk esa gS] rkjksa ds e/; fLFkr i1
fcUnq ij pqEcdh; {ks=k dh rhozrk 10T gSA ;fn i2 dh fn'kk ifjofrZr dj nh tk;s rks {ks=k 30T gks tkrk gSA vuqikr i 2 dk eku gksxk (A) 4 12.
(B) 3
(C*) 2
(D) 1
Consider a long, straight wire of cross-section area A carrying a current i.Let there be n free electrons per unit volume. An observer places himself on a trolley moving in the direction opposite to the current with a speed = (i/nAe) and separated from the wore bu a distance r. The magnetic field seen by the observer is very nearly A vuqiLz Fk dkV {ks=kQy ds ,d yEcs rFkk lh/ks rkj ls i /kkjk izokfgr gks jgh gSA ekuk fd blds ,dkad vk;ru esa n eqDr bysDVªkWu gSA Vªky W h ij cSBk gqvk ,d iz{s kd rkj ls r nwjh ij /kkjk ds foifjr fn'kk esa = (i/nAe) pky ls xfr'khy gSA iz{s kd }kjk izfs {kr pqEcdh; {ks=k yxHkx gksxk (A*)
0i 2 r
(B) zero
(C)
0i r
(D)
20i r
Objective - II
1.
The magnetic field at the origin due to a current element i d placed at a position r is ewy fcUnq ij fLFkr ,d /kkjk vo;o i d ds dkj.k fLFkfr r ij pqEcdh; {ks=k -
0i dl xr 0i dl xr (A) (B) 4 r 3 4 r 3
(C*)
0i rxdl 4 r 3
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(D*)
0i dl xr 4 r 3 Page # 4
Chapter # 35
Magnetic Field due to A current
2.
Consider three quantities x = E/B, y = 1/oo and z = / CR. Here, l is the length of a wire, C is a capacitance and R is a resistance. All other symbols have standard meanings (A*) x, y have the same dimensions. (B*) y, z have the same dimensions (C*) z, x have the same dimensions (D) None of the three pairs have the same dimensions rhu jkf'k;ksa x = E/B, y = 1/oo rFkk z = / CR ij fopkj dhft;sA buesa rkj dh yEckbZ] C /kkfjrk rFkk R izfrjks/k gS] vU; ladrs ksa ds lkekU; vFkZ gS (A*) x, y dh foek,¡ ,d leku gSA (B*) y, z dh foek,¡ ,d leku gSA (C*) z, x dh foek,¡ ,d leku gSA (D) rhuksa esa ls fdlh Hkh ;qXe dh foek,¡ ,d leku ugha gSA
3.
A long, straight wire carries a current along the Z-axis. One can find two points in the X-Y plane such that (A) the magnetic fields are equal (B*) the directions of the magnetic fields are the same (C*) the magnitudes of the magnetic fields are equal (D*) the field at one point is opposite to that at the other point ,d yEcs ,oa lh/ks rkj esa Z-v{k ds vuqfn'k /kkjk izokfgr gks jgh gSA x - y ry esa fLFkr nks fcUnqvksa ds fy;s dksbZ O;fDr izfs {kr dj ldrk gS] fd (A) pqEcdh; {ks=k ,d leku gSA (B*) pqEcdh; {ks=kksa dh fn'kk,¡ ,d leku gSA (C*) pqEcdh; {ks=kksa ds ifjek.k ,d leku gSA (D*) ,d fcUnq ij pqEcdh; {ks=k dh fn'kk nwljs fcUnq ij {ks=k dh fn'kk ls foifjr gSA
4.
A long, straight wire of radius R carries a current distrobuted uniformly over its cross-section. The magnitude of the magnetic field is (A) maximum at the axis of the wire (B*) minimum at the axis of the wire (C*) maximum at the surface of the wire (D) minimum at the surface of the wire. R f=kT;k ds ,d yEcs ,oa lh/ks rkj ls gksus okyk /kkjk izokg lEiw.kZ vuqiLz Fk dkV esa ,d leku forfjr gSA pqEcdh; {ks=k dk ifjek.k gS (A) rkj dh v{k ij vf/kdre (B*) rkj dh v{k ij U;wure (C*) rkj dh lrg ij vf/kdre (D) rkj dh lrg ij U;wure
5.
A hollow tube is carrying an electric current along its length distributed uniformly over its surface. The magnetic field (A) increases linearly from the axis to the surface (B*) is constant inside the tube (C*) is zero at the axis (D) is zero just outside the tube
,d [kks[kyh ufydk dh yEckbZ ds vuqfn'k gksus okyk /kkjk izokg] bldh lrg ij ,d leku :i ls forfjr gSA pqEcdh; {ks=k (A) v{k ls lrg dh vksj jsf[kd :i ls c<+rk gSA (B*) ufydk ds vUnj fu;r jgrk gSA (C*) v{k ij 'kwU; jgrk gSA (D) ufydk ds rqjUr ckgj 'kwU; gksrk gSA 6.
In a coaxial, straight cable, the central conductor and the outer conductor carry equal currents in opposite directions. The magnetic field is zero. (A*) outside the cable (B) inside the inner conductor (C) inside the outer conductor (D) in between the two conductors.
,d lek{kh;] lh/kh dscy esa dsUnzh; pkyd rFkk cká pkyd ls leku /kkjk,¡ ijLij foijhr fn'kkvksa esa izokfgr gks jgh gSA pqEcdh; {ks=k 'kwU; gksxk (A*) dscy ds ckgj (B) vkarfjd pkyd ds vUnj (C) cká pkyd ds vUnj (D) nksuksa pkydksa ds chp esa 7.
A steady electric current is flowing through a cylindrical conductor. (A) the electric field at the axis of the conductor is zero (B*) the magnetic field at the axis of the conductor is zero (C*) the electric field in the vicinity of the conductor is zero (D) the magnetic field in the vidinity of the conductor is zero ,d csyukdkj pkyd ls LFkk;h fo|qr/kkjk izokfgr gks jgh gS (A) pkyd dh v{k ij fo|qr {ks=k 'kwU; gksxkA (B*) pkyd dh v{k ij pqEcdh; {ks=k 'kwU; gksxkA (C*) pkyd ds lehi fo|qr {ks=k 'kwU; gksxkA (D) pkyd ds lehi pqEcdh; {ks=k 'kwU; gksxkA manishkumarphysics.in
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Chapter # 35
1.
Magnetic Field due to A current
WORKED OUT EXAMPLES
Two long wires a and b, carrying equal currents of 10.0 A, are placed parallel to each other with a separation of 4.00 cm between them as shown in figure. Find the magnetic field B at each of the points P, Q and R. 2.00 cm P
Sol.
2.00 cm
2.00 cm
2.00 cm
Q
a
R
b
The magnetic field at P due to the wire has magnitude T m 4 10 7 10 A 0i A B1 = – 2d 2 2 10 2 m = 1.00 × 10–4 T. Its direction will be perpendicular to the line shown and will point downward in the figure. The field at this point due to the other wire has magnitude Similarly, the resultant magnetic field at R will be = 1.33 × 10–4 T along the direction pointing upward in the figure. The magnetic field at point Q due to the two wires will have equal magnitudes but opposite directions and hence the resultant field will be zero.
2.
Two parallel wires P and Q placed at a separation d = 6 cm carry electric currents i1 = 5A and i2 = 2A in opposite directions as shown in figure (a). Find the point on the line PQ where the resultant magnetic field is zero. i1 P
Sol.
i1
i2 d (a)
P
Q
i2 d (b)
Q x
R
At the desired point, the magnetic fields due to the two wires must have equal magnitude but opposite directions. The point should be either to the left of P or to the right of Q. As the wire Q has smaller current, the point should be closer to Q. Let this point R be at a distance x from Q (figure (b)). The magnetic field at R due to the current i1 will have magnitude
B1
0i1 2(d x )
and will be directed downward in the plane of the figure. The field at the same point due to the current i2 will be B2
0i 2 2x
directed upward in the plane of the figure. If the resultant field at R is zero, we should have B1 = B2, so that i1 i 2 d x x
giving,
3.
i2 d x = i i 1 2
=
(2 A )(6 cm) 4cm (3 A )
Two long, straight wires a and b are 2.0 m apart, perpendicular to the plane of the paper as shown in figure. The wire a carries a current of 9.6 A directed into the plane of the figure. The magnetic field at the point p at
10 metre from the wire b is zero. Find (a) the magnitude and direction of the current in b, (b) 11 the magnitude of the magnetic field B at the point s and (c) the force per unit length on the wire b. a distance of
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Chapter # 35
Magnetic Field due to A current a 1.6 m 2.0m
s 1.2 m
b 10/11m
P
Sol.
(a) For the magnetic field at p to be zero, the current in the wire b should be coming out of the plane of the figure so that the fields due to a and b may be opposite at p. The magnitude of these fields should be equal, so that 0 (9.6 A ) 0i 10 10 2 2 m 2 m 11 11 or,
i = 3.0 A
(b)
(ab)2 = 4m2 (as)2 = 2.56 m 2 (bs)2 = 1.44 m 2
and
so that (ab) 2 = (as) 2 + (bs) 2 and angle asb = 90º. The field at s due to the wire a
0 ( 9.6 A ) 0 A = 2 1.6 m = 2 × 6 m and that due to the wire b =
0 3A 0 A = × 2.5 2 1.2 m 2 m
These fields are at 90º to each other so that their resultant will have a magnitude 2
A A 0 6 0 2. 5 2 m 2 m
=
0 2
2
A 36 6.25 m
= 2 × 10–6 T. (c)
The force per unit length on the wire b =
(9.6 A )(3 A ) 0 i1i 2 7 T m × = 2 10 2.0 m A 2d
= 2.9 × 10–6 N 4.
A current of 2.00 A exists in a square loop of edge 10.0 cm. Find the magnetic field B at the centre of the square loop.
Sol.
The magnetic fields at the centre due to the four sides will be equal in magnitude and direction. The field due to one side will be B1 =
0 ia 2d a 2 4d2
.
Here, a = 10 cm and d = a/2 = 5 cm. Thus,
B1
10 cm 0 (2 A ) 2 2 2(5 cm) (10 cm) 4(5 cm)
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Chapter # 35
Magnetic Field due to A current = 2 × 10–7
1 T m × 2A × 5 2cm A
= 5.66 × 10–6 T. Hence, the net field at the centre of the loop will be 4 × 5.66 × 10–6 T = 22.6 × 10–6 T. 5.
Figure shows a square loop made from a uniform wire. Find the magnetic field at the centre of the square if a battery is connected between the points A and C. C
D
B
A
Sol.
The current will be equally divided at A. The fields at the centre due to the currents in the wires AB and DC will be equal in magnitude and opposite in direction. The resultant of these two fields will be zero. Similarly, the resultant of the fields due to the wires AD and BC will be zero. Hence, the net field at the centre will be zero.
6.
Two long wires, carrying currents i1 and i2, are placed perpendicular to each other in such a way that they just avoid a contact. Find the magnetic force on a small length dl of the second wire situated at a distance l from the first wire.
i1 i2
Sol.
The situation is shown in figure. The magnetic field at the site of dl, due to the first wire is , 0i1 2 This field is perpendicular to the plane of the figure going into it. The magnetic force on the length d is, dF = i2 d B sin 90º B
0i1i 2 d 2 This force is parallel to the current i1.
=
7.
Figure shows a part of an electric circuit. ABCD is a rectangular loop made of uniform wire. The length AD = BC = 1 cm. The sides AB and DC are long as compared to the other two sides. Find the magnetic force A
B
D
C
A
Sol.
per unit length acting on the figure DC due to the wire AB if the ammeter reads 10 A. By symmetry, each of the wires AB and DC carries a current of 5 A. As the separation between them is 1 cm, the magnetic force per unit length of DC is dF 0i1i 2 d 2d
1 T m 2 10 (5 A ) (5 A ) A = 1 10 2 m
= 5 × 10–4 T–A = 5 × 10–4 N-m.
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Chapter # 35 Magnetic Field due to A current 8. Figure shows a current loop having two circular arcs joined by two radial lines. Find the magnetic field B at the centre Q. i D
C B
A
a b
O
Sol.
As the point O is on the line AD, the magnetic field at O due to AD is zero. Similarly, the field at O due to BC is also zero. The field at the centre of a circular current loop is given by B =
0i . The field due to the circular 2a
are BA will be i B1 = 0 2a 2b going into the plane of the figure. The resultant field at O is
B = B1 – B2 =
0i(b a) 4ab
coming out of the plane. 9.
Find the magnetic field at the point P in figure. The curved portion is a semicircle and the straight wire are long. i
P
d i
Sol.
The magnetic field at P due to any current element in the figure is perpendicular to the plane of the figure and coming out of it. The field due to the upper straight wire is B1 =
1 × 2
i 0i = 0 . 2 d d 2 2
Same is the field B2 due to the lower straight wire. The field due to the semicircle of radius (d/2) is B2 =
1 × 2
0i i = 0 . 2d d 2 2
The net field is B = B1 + B2 = 10. Sol.
2 0i 1 . 2d
The magnetic field B due to a current-carrying circular loop of radius 12 cm at its centre is 0.50 × 10–4 T. Find the magnetic field due to this loop at a point on the axis at a distance of 5.0 cm from the centre. The magnetic field at the centre of a circular loop is 0i 2a and that at an axial point is
B0 =
B=
0ia2 2(a 2 x 2 )3 / 2
B a3 Thus, B = 2 0 (a x 2 ) 3 / 2 manishkumarphysics.in
Page # 9
Chapter # 35 or,
Magnetic Field due to A current B = (0.50 × 10–4 T) ×
(12 cm)3 (144 cm 2 25 cm 2 )3 / 2
= 3.9 × 10–5 T. 11.
Consider a coaxial cable which consists of an inner wire of radius a surrounded by an outer shell of inner and outer radii b and c respectively. The inner wire carries an electric current i0 and the outer shell carries an equal current in opposite direction. Find the magnetic field at a distance x from the axis where (a) x < a, (b) a < x < b (c) b < x < c and (d) x > c. Assume that the current density is uniform in the inner wire and also uniform in the outer shell.
Sol.
(a)
(b)
(c)
(d)
A cross-section of the cable is shown in figure. Draw a circle of radius x with the centre at the axis of the cable. The parts a, b, c and d of the figure correspond to the four parts of the problem. By symmetry, the magnetic field at each point of a circle will have the same magnitude and will be tangential to it. The circulation of B along this circle is, therefore, B.d B2x
in each of the four parts of the figure. (a) The current enclosed within the circle in part is
i0
i0
. x2 =
x2 .
a a B.d 0i gives 2
2
Ampere’s law
B.2x =
0 i0 x 3
B 2x = 0 i0
or, B =
0 i0 x
. 2a 2 a The direction will be along the tangent to the circle. (b) The current enclosed within the circle in part b is i0 so that 2
or,
B=
0 i0 x
. 2a 2 (c) The area of cross-section of the outer shell is c2 – b2. The area of cross-section of the outer shell with in the circle in part c of the figure is x2 – b2. The area of cross-section of the outer shell within the circle in part c of the figure is x2 – b2.
i0 ( x 2 b 2 )
Thus, the current through this part is
(c 2 b 2 )
. This is in the opposite direction to the current i0 in the inner
wire. Thus, the net current enclosed by the circle is i0 =
i0 ( x 2 b 2 ) c b 2
2
=
i0 (c 2 x 2 ) c 2 b2
.
Form Ampere’s law, B 2x =
or,
B=
0 i0 (c 2 x 2 ) c 2 b2
0 i 0 (c 2 x 2 ) 2x(c 2 b 2 ) manishkumarphysics.in
Page # 10
Chapter # 35 Magnetic Field due to A current (d) The net current enclosed by the circle in part d of the figure is zero and hence B 2x = 0 or, B = 0-. 12.
Figure shows a cross-section of a large metal sheet carrying an electric current along its surface. The current in a strip of width dl is Kdl where K is a constant. Find the magnetic field at a point P at a distance x from the metal sheet.
P x
Sol.
Consider two strips A and C of the sheet situated symmetrically on the two sides of P (figure). The magnetic field at P due to the strip A is B0 perpendicular to AP and that due to the strip C is BC perpendicular to CP. The resultant of these two is parallel to the width AC of the sheet. The field due to the whole sheet will also be in this direction. Suppose this field has magnitude B. B P
Ba P
A dl
x
x
Bc
x
C
O
B
dl
The field on the opposite side of the sheet at the same distance will also be B but in opposite direction. Applying Ampere’s law to the rectangle shown in figure. 2B = m0 K
1 K. 2 0 Note that it is independent of x. or,
13.
Sol.
B=
Consider the situation described in the previous example. A particle of mass m having a charge q is placed at a distance d from the metal sheet and is projected towards it. Find the maximum velocity of projection for which the particle does not hit the sheet. As the magnetic field is uniform and the particle is projected in a direction perpendicular to the field, it will describe a circular path. The particle will not hit the metal sheet. If the radius of this circle is smaller than d. For the maximum velocity, the radius is just equal to d. Thus. qvB =
14.
mv 2 d
0K mv 2 2 d
or,
qv
or,
v
0 qKd 2m
Three identical long solenoids P, Q and R are connected to each other as shown in figure. If the magnetic field at the centre of P is 2.0 T, what would be the field at the centre of Q? Assume that the field due to any solenoid is confined within the volume of that solenoid only.
Q P
R Sol.
As the solenoids are identical, the currents in Q and R will be the same and will be half the current in P. The magnetic field within a solenoid is given by B = 0ni. Hence the field in Q will be equal to the field in R and will be half the field in P i.e., will be 1.0 T. manishkumarphysics.in
Page # 11
Chapter # 35 15.
Magnetic Field due to A current
A long, straight wire carries a current i. A particle having a positive charge q and mass m kept at a distance x0 from the wire is projected towards it with a speed v. Find the minimum separation between the wire and the particle
Y i x0 Sol.
x
P
Let the particle be initially at P (figure). Take the wire as the Y-axis and the foot of perpendicular from P to the wire as the origin. Take the line OP as the X-axis. We have, OP = x0. The magnetic field B at any point to the right of the wire is along the negative Z-axis. The magnetic force on the particle is, therefore, in the X–Y plane. As there is no initial velocity along the Z-axis, the motion will be in the X–Y plane. Also, its speed remains unchanged. As the magnetic field is not uniform, the particle does not go along a circle. The force at time t is F qv B
i = q( i vx + j vy) × 0 k 2x i qi = j qv x 0 0 – i qvy 0 . 2x 2m Thus
ax =
vy 0 qi v y Fx =– =– 2m x m x
where
=
0 qi . 2m
Also,
ax =
As, giving From (i), (ii) and (iii),
dv x dv x dx v x dv x = dt dx dt dx 2 2 2 vx + vy = v , vx dvx = – vy dvy.
v y dv y
dx
....(i)
....(ii) .....(iii)
v y x
dx dv y x Initially x = x0 and vy = 0. At minimum separation from the wire, vx = 0 so that vy = –v. or,
x
Thus
x0
dv y
0
x v ln x = – 0
or,
–v/
or,
1.
v
dx x
x = x0 e
= x0 e
2 mv 0 qi
.
EXERCISE
0i Using the formulae F qv B and B = ,. show that the SI unis of the magnetic field B and thye 2r permeability constant 0 may be written as N/A-m and N/A2 respectively.
i
lw=k F qv B rFkk B = 0 dk mi;ksx djds ;g iznf'kZr dhft;s fd pqEcdh; {ks=k B rFkk ikjxE;rk fu;rkad 0 dh 2r bdkbZ;k¡ Øe'k% N/A-m rFkk N/A2 fy[kh tk ldrh gSA manishkumarphysics.in
Page # 12
Chapter # 35 2.
Magnetic Field due to A current
A current of 10 A is established in a long wire along the positive Z-axis. Find the magnetic field B at the point (1m, 0, 0). /kukRed Z-v{k ds vuqfn'k fLFkr ,d yEcs rkj esa 10 A /kkjk izokfgr dh tk jgh gSA fcUnq (1m, 0, 0) ij pqEcdh; {ks=k B
Kkr dhft;sA Ans. 3.
2T along the positive Y-axis
A copper wire of diameter 1.6 mm carries a current of 20 A. Find the maximum magnitude of the magnetic field B due to this current. 1.6 feeh O;kl okys ,d rkacs ds rkj ls 20 A /kkjk izokfgr gks jgh gSA bl /kkjk ds dkj.k mRiUu pqEcdh; {ks=k B dk
vf/kdre ifjek.k Kkr dhft;sA Ans.
5.0 m T
4.
A transmission wire carries a current of 100 A. What would be the magnetic field B at a point on the road if the wire is 8 m above the road ? ,d lapj.k rkj esa 100 A /kkjk izokfgr gks jgh gSA ;fn rkj lM+d ls 8 eh- Åij gS rks lM+d ij fLFkr fdlh fcUnq ij pqEcdh; {ks=k dh rhozrk B fdruh gksxh\
5.
A long, straight wire carrying a current of 1.0 A is placed horizontally in a uniform magnetic field B = 1.0 × 10–5 T pointing vertically upward (figure). Find the magnitude of the resutlant magnetic field at the points P and Q, both situated at a distance of 2.0 cm from the wire in the same horizontal plane. Å¡/okZ/kj Åij dh vksj bafxr le:i pqEcdh; {ks=k B = 1.0 × 10–5 T esa {kSfrt j[ks gq, ,d yEcs ,oa lh/ks rkj ls 1.0 A /kkjk izokfgr gks jgh gS ¼fp=k½A fcUnqvksa P rFkk Q, ij ifj.kkeh pqEcdh; {ks=k dk eku Kkr dhft;sA nksuksa fcUnq {kSfrt ry esa rkj ls 2.0 lseh nwj fLFkr gSA
6.
A long, straight wire of radius r carries a current i and is placed horizontally in a uniform magnetic field B pointing vertically upward. The current is uniformly distributed over its cross-section. (a) At what points will the resultant magnetic field have maximum magnitude? What will be that maximum magnitude? (b) What will be the minimum magnitude of the resultant magnetic field? Å/okZ/kj Åij dh vksj bafxr ,d le:i pqEcdh; {ks=k B esa ,d yEck ,oa lh/kk r f=kT;k dk {kSfrt rkj fLFkr gSA rkj dh f=kT;k r gSA rkj esa çokfgr /kkjk i lEiw.kZ vuqiLz Fk dkV esa ,d leku :i ls forfjr gSA (a) fdu fcUnqvksa ij ifj.kkeh pqEcdh; {ks=k dk eku vf/kdre gksxk \ vf/kdre ifjek.k fdruk gksxk\ (b) ifj.kkeh pqEcdh; {ks=k dk U;wure eku fdruk gksxk\ [5 min.] [M.Bank(07-08)_HCV_Ch.35_Ex._6] Ans.
(a) looking along the current, at the leftmost points on the wire’s surface, B +
0i 2r
0i 0i 0i ,B– if r > 2B 2r 2B A long vertical wire carying a current of 30 A is placed in an external, uniform magnetic field of 4.0 × 10–4 T parallel to the current. Find the magnnitude of the resultant magnetic field at a point 2.0 cm away from the wire. ,d yEck ,oa lh/ks rkj ls 30 A /kkjk izokfgr gks jgh gSA ;g rkj ,d cká le:i pqEcdh; {ks=k esa j[kk gqvk gSA ftldh rhozrk 4.0 × 10–4 T rFkk fn'kk /kkjk ds lekUrj gSA rkj ls 2.0 lseh nwjh ij ifj.kkeh pqEcdh; {ks=k ifj.kke Kkr dhft;sA
(b) zero if r
7.
8.
A long, vertical wire carrying a current of 10 A in the upward direction is placed in a region where a horizontal magnetic field of magnitude 2.0 × 10–3 T exists from south to north. Find the point where the resultant magnetic field is zero. nf{k.k ls mÙkj dh vksj fn"V le:i ,oa {kSfrt pqEcdh; {ks=k dh rhozrk 2.0 × 10–3 T gS] blesa fLFkr ,d yEcs ,oa lh/ks rkj esa Å/okZ/kj Åij dh vksj 10 A /kkjk izokfgr gks jgh gSA og fcUnq Kkr dhft;s] tgk¡ ifj.kkeh pqEcdh; {ks=k 'kwU; gSA Ans. 5 × 10–4 T
9.
Figure shows two parallel wires separated by a distance of 4.0 cm and carrying equal currents of 10 A along opposite directions. Find the magnitude of the magnetic field B at the point A1, A2, A2 and A4. fp=k esa nks lekukUrj rkjksa dks iznf'kZr fd;k x;k gSA ftuds e/; 4.0 lseh nwjh gSA bu rkjksa esa 10 A dh leku /kkjk,¡ ijLij foifjr fn'kkvksa esa izokfgr gks jgh gSA fcUnqvksa A1, A2, A2 rFkk A4 ij pqEcdh; {ks=k B dk ifjek.k Kkr dhft;sA
manishkumarphysics.in
Page # 13
Chapter # 35
Ans.
Magnetic Field due to A current
5 × 10–4 T
10.
Two parallel wires carry equal currents of 10 A along the same direction and are separated by a distance of 2.0 cm. Find the magnetic field at a point which is 2.0 cm away from each of these wires. nks lekukUrj rkjksa ds e/; 2.0 lseh nwj gS] rFkk buesa 10 A dh leku /kkjk,¡ ,d gh fn'kk esa izokfgr gks jgh gSA bu rkjksa ls izR;sd ls 2.0 lseh nwjh ij fLFkr fcUnq ij pqEcdh; {ks=k Kkr dhft;sA Ans. (a) 0.67 × 10–4 T (b) 2.7 × 10–4 T (c) 2.0 × 10–4 T (d) 1.0 × 1.0–4 T
11.
Two long, straight wires, each carrying a current of 5 A, are placed along the X-axis and Y-axis respectively. The currents point along the positive direction of the axes. Find the magnetic fields at the points (a) (1 m , 1m), (b) (–1 m, 1m), (c) (–1 m, –1m) and (d) (1m, – 1m). nks yEcs ,oa lh/ks rkjksa esa izR;sd ls 5 A /kkjk izokfgr gks jgh gSA ;g rkj Øe'k% X-v{k rFkk Y-v{kksa ds vuqfn'k j[ks gq, gSAa /kkjkvksa dh fn'kk,¡ /kukRed v{kksa ds vuqfn'k gSA fcUnqvksa (a) (1eh , 1eh), (b) (–1eh, 1 eh), (c) (–1 eh, –1eh) vkSj (d) (1eh, – 1eh) Ans. (a) zero (b) 2mT along the Z-axis (c) zero and (d) 2T along the negative Z-axis
12.
Four long straight wires, each carrying a current of 5.0 A, are placed in a placed in a plane as shown in figure. The points of intersection from square of side 5.0 cm. (a) Find the magnetic field at the centre P of the square. (b) Q1, Q2, Q3 and Q4 are points situated on the diagonals of the square and at a distance from P that is equal to the length of the diagonal of the square. Find the magnetic fields at these points. pkj yEcs ,oa lh/ks rkjksa esa izR;sd ls 5.0 A /kkjk izokfgr gks jgh gS] budks fp=kkuqlkj j[kk x;k gSA rkjksa ds dVku fcUnq 5.0 lseh Hkqtk dk oxZ fufeZr djrs gSAa (a) oxZ ds dsUnz P ij pqEcdh; {ks=k Kkr dhft;sA (b) Q1, Q2, Q3 rFkk Q4 fcUnq P ls fod.kZ
dh yEckbZ ds cjkcj nwjh ij fLFkr gSA bu fcUnqvksa ij pqEcdh; {ks=k Kkr dhft;sA
Ans. 13.
(a) zero (b) Q1 : 1.1 × 10–4 T,
, Q2 : zero, Q3 : 1.1 × 10–4 T,
× , and Q
4
: zero
Figures shows a long wire bent at the middle to form a right angle. Show that the magnitudes of the magnetic fields at the points P, Q, R and S are equal and find this magnitude. fp=k esa ,d yEck rkj iznf'kZr fd;k x;k gSA ftls e/; esa ledks.k ij eksMk+ x;k gSA iznf'kZr dhft;s fd fcUnqvksa P, Q, R rFkk S ij pqEcdh; {ks=kksa ds ifjek.k leku gS rFkk bl {ks=k dk eku Kkr dhft;sA
0i 4d Consider a straight piece of length x of a wire carrying a current i. Let P be a point on the perpendicular bisector of the piece, situated at a distance d from its middle point. Show that for d > > x, the magnetic field at P varies as 1/d2 whereas for d < < x, it varies as 1/d. rkj ds x yEckbZ ds ,d lh/ks VqdM+s ij fopkj dhft;sA ftlls i /kkjk izokfgrgks jgh gSA ekuk fd rkj ds yEcv/kZd ij]
Ans. 14.
manishkumarphysics.in
Page # 14
Chapter # 35
Magnetic Field due to A current
e/; fcUnq ls d nwjh ij ,d fcUnq ij ,d fcUnq P ij pqEcdh; {ks=k d > > x ds fy;s 1/d2 ds vuqlkj ifjofrZr gksrk gSA tcfd d < < x ds fy, 1/d ds vuqlkj ifjofrZr gksrk gSA 15.
Consider a 10 cm long piece of a wire which carries a current of 10 A. Find the magnitude of the magnetic field due to the piece at a point which makes an equilateral triangle with the ends of the piece. 10 lseh yEcs rkj ds VqdM+s ij fopkj dhft;sA ftlls 10A /kkjk izokfgr gks jgh gSA blds fljksa ls leckgq f=kHkqt cukus okys
fcUnq ij bl VqdM+s ds dkj.k pqEcdh; {ks=k dk ifjek.k Kkr dhft;sA Ans. 16.
11.5 T
A long, straight wire carries a current i. Let B1 be the magnetic field at a poit P at a distance d from the wire. Consider a section of length l of this wire such that the point P lies on a perpendicular bisector of the section. Let B2 be the magnetic field at this point due to this section only. Find the value of d/ so that B2 differs from B1 by 1%. ,d yEcs ,oa lh/ks rkj ls i /kkjk izokfgr gks jgh gSA ekuk fd rkj ls d nwjh ij fLFkr fcUnq P ij pqEcdh; {ks=k B1 gSA bl rkj ds yEckbZ ds ,d Hkkx ij fopkj dhft;s tks ,d izdkj fLFkr gS fd bl Hkkx ds yEc v/kZd ij fcUnq P fLFkr gSA ekuk fd dsoy bl Hkkx ds dkj.k fcUnq P ij pqEcdh; {ks=k B2 gSA d/ dk eku Kkr dhft;sA ftlds fy;s B2 dk B1 ls vUrj 1%
gksA 17.
Figure shows a square loop ABCD with edge-length . The resistance of the wire ABC is r and that of ADC is 2r. Find the magnetic field B at the centre of the loop assuming uniform wires. fp=k esa ,d oxkZdkj ywi ABCD iznf'kZr gSA ftldh Hkqtk dh yEckbZ a gSA rkj ABC dk izfrjks/k r rFkk rkj ADC dk izfrjks/k 2r gSA rkjksa dk le:i ekurs gq, oxZ ds dsUnz ij pqEcdh; {ks=k Kkr dhft;sA
2 0i , 3d Figure shows a square loop of edge a made of a uniform wire. A current i enters the loop at the point A and leaves it at the point C. Find the magnetic field at the point P which is on the perpendicular bisector of AB at a distance a/4 from it. fp=k esa iznf'kZr oxkZdkj ywi le:i rkj dk cuk gqvk gS rFkk bldh Hkqtk a gSA ywi ds A fcUnq ij i /kkjk izfo"V gks jgh gS rFkk ;g /kkjk fcUnq C ls ckgj fudy jgh gSA AB ds yEcv/kZd ij blls a/4 nwjh ij fLFkr fcUnq P ij pqEcdh; {ks=k Kkr Ans.
18.
dhft;sA
D
C i P a/4
i
Ans. 19.
2 0 i a
A
B
1 1 , 5 3 13
Consider the situation described in the previous problem. Suppose the current i enters the loop at the point A and leaves it at the point B. Find the magnetic field at the centre of the loop. fiNys iz'u esa of.kZr fLFkr ij fopkj dhft;sA ekukfd A fcUnq ij i /kkjk ywi esa izfo"V gksrh gS rFkk B fcUnq ls ckgj fudyrh
gSA ywi ds dsUnz ij pqEcdh; {ks=k Kkr dhft;sA Ans. 20.
zero
The wire ABC shown in figure forms an equilateral triangle. Find the magnetic field B at the centre O of the triangle assuming the wire to be uniform. fp=k esa iznf'kZr rkj ABC ,d leckgq f=kHkqt cukrk gSA rkj dks le:i ekurs gq, f=kHkqt ds dsUnz O ij pqEcdh; {ks=k B Kkr
dhft;sA
manishkumarphysics.in
Page # 15
Chapter # 35
Magnetic Field due to A current
A
O B
C
21.
A wire of length isbent in the form of an equilateral triangle and carries an electric current i. (a) Find the magnetic field B at the centre. (b) If the wire is bent in the form of a square, what would be the value of B at the centre ? yEckbZ dk ,d rkj leckgq f=kHkqt dh vkÑfr esa eksMk+ x;k gS rFkk blls i fo|qr /kkjk izokfgr gks jgh gSA (a) dsUnz ij pqEcdh; {ks=k B Kkr dhft;sA (b) ;fn rkj dks oxkZdkj vkÑfr esa eksM+k x;k gksrk rks dsUnz ij B fdruk gksrk?
22.
A long wire carrying a current i is bent to form a plane angle . Find the magnetic field B at a point on the bisector of this angle situated at a distance x from the vertex. ,d yEcs rkj ls i /kkjk izokfgr gks jgh gS] bldks bl izdkj eksM+k x;k gS fd ;g lery esa dks.k cukrk gSA bl dks.k ds v/kZd ij rFkk 'kh"kZ ls x nwjh ij fLFkr fcUnq ij pqEcdh; {ks=k Kkr dhft;sA Ans.
23.
0i a cot 2x 4
Find the magnetic field B at the centre of a rectangular loop of length l and width b, carrying a current i. yEckbZ rFkk b pkSM+kbZ ds vk;rkdkj ywi ls i /kkjk izokfgr gks jgh gS] blds dsUnz ij pqEcdh; {ks=k B Kkr dhft;sA
2 0i l 2 b 2 2lb A regular polygon of n sides is formed by bending a wire of total length 2r which carries a current i. (a) Find the magnetic field B at the centre of the polygon. (b) By letting n , deduce the expression for the magnetic field at the centre of a circular current. ,d rkj dh dqy yEckbZ 2r gS rFkk blls i /kkjk izokfgr gks jgh gSA bl rkj dks eksMd + j n ,d leku Hkqtkvksa okyk le cgqHkqt cuk;k tkrk gSA (a) cgqHkqt ds dsUnz ij pqEcdh; {ks=k Kkr dhft;sA (b) n , ekurs gq, ,d o`Ùkkdkj /kkjk ds dsUnz ij Ans.
24.
pqEcdh; {ks=k ds fy;s O;atd O;qRiUu dhft;sA Ans. 25.
0 in 2 sin tan n n (a) 2 2r
(b)
Each of the batteries shown in figuer has an emf equal to 5 V. Show that the magnetic field B at the point P is zero for any set of values of the resistances. fp=k esa iznf'kZr izR;sd cSVjh dk fo-ok-cy 5 V gSA iznf'kZr dhft;s fd izfrjks/kksa ds lewg ds fdUgh Hkh ekuksa ds fy;s fcUnq P ij pqEcdh; {ks=k B dk eku 'kwU; gh jgsxkA
P
Ans. 26.
A straight, long wire carries a current of 20 A. Another wire carrying equal current is placed parallel to it. If the force acting on a length of 10 cm of the second wire is 2.0 × 10–5 N, what is the separation between them ? ,d lh/ks ,oa yEcs rkj ls 20 A /kkjk izokfgr gks jgh gSA ,d vU; rkj blds lekukUrj j[kk x;k gSA ftlls leku /kkjk izokfgr gks jgh gSA ;fn nwljs rkj dh 10 lseh yEckbZ ij 2.0 × 10–5 N cy yx jgk gks rks nksuksa ds e/; fdruh nwjh gS\ Ans. 40 cm
27.
Three coplanar parallel wires each carrying a current of 10 A along the same direction, are placed with a separation 5.0 cm between the consecutive ones. Find the magnitude of the magnetic force per unit length manishkumarphysics.in
Page # 16
Chapter # 35
Magnetic Field due to A current
acting on the wires.
,d lery esa fLFkr rhu lekukUrj rkjksa esa izR;sd ls 10 A /kkjk ,d gh fn'kk esa izokfgr gks jgh gSA buds e/; Øekxr :i ls 5.0 lseh nwjh gSA rkjksa dh ,dkad yEckbZ ij yxus okys cy dk ifjek.k Kkr dhft;sA Ans. 28.
zero on the middle wire and 6.0 × 10–4 N towards the middle wire on each of the rest two
Two parallel wires separated by a distance of 10 cm carry currents of 10 A and 40 A along the same direction. Where should a third current be placed so that it experiences no magnetic force? 10 lseh nwjh ij fLFkr nks lekukUrj rkjksa ls 10 A rFkk 40 A /kkjk,¡ ,d gh fn'kk esa izokfgr gks jgh gSA ,d rhljs /kkjkokgh
rkj dks fdl LFkku ij j[kk tk;s fd ;g dksbZ pqEcdh; cy vuqHko ugha djs\ Ans. 29.
2 cm from the 10 A current and 8 cm from the other
Figure shows a part of an electric circuit. The wires AB, CD and EF are long and have identical resistances. The separation between the neighbouring wires si 1.0 cm. The wires AE and BF have negligible resistance and the ammeter reads 30 A. Calculate the magnetic force per unit length of AB and CD. fp=k esa fdlh fo|qr ifjiFk dk ,d Hkkx iznf'kZr gSA rkj AB, CD ,oa EF yEcs gS rFkk buds izfrjks/k ,d leku gSA lehi okys nks rkjksa ds e/; 1.0 lseh nwjh gSA rkjksa AE rFkk BF ds izfrjks/k ux.; gS vkSj vehVj dk ikB~;kad 30 A gSA AB rFkk CD dh bdkbZ yEckbZ ij pqEcdh; cy ds ifjek.k dh x.kuk dhft;sA
A
Ans.
A
B
C
D
E
F
3 × 10–3 N/m, downward zero
30.
A long, straight wire is fixed horizontally and carries a current of 50.0 A, A second wire having linear mass density 1.0 × 10–4 kg/m is placed parallel to and directly above this wire at a separation of 5.0 mm. What current should this second wire carry such that the magnetic repulsion can balance its weight ?
31.
A square loop PQRS carrying a current of 6.0 A is placed near a long wire carrying 10 A as shown in figure (a) Show that the magnetic force acting on the part PQ is equal and opposite to that on the part RS. (b) Find the magnetic force on the square loop. ,d yEcs rkj ls 10 A /kkjk izokfgr gks jgh gS] fp=kkuqlkj blds lehi fLFkr ,d oxkZdkj ywi PQRS ls 6.0 A /kkjk izokfgr gks jgh gSA (a) iznf'kZr dhft;s fd Hkkx PQ ij pqEcdh; cy] Hkkx RS ij cy ds cjkcj rFkk foifjr gSA (b) oxkZdkj ywi
ij pqEcdh; cy Kkr dhft;sA
2cm S
R P
P
Q
1cm
Ans.
(b) 1.6 × 10–6 N towards right.
32.
A circular loop of one turn carries a current of 5.00 A. If the magnetic field B at the centre is 0.200 mT, find the radius of the loop.
33.
A current carrying circular coil of 100 turns and rdius 5.0 cm produces a magnetic field of 6.0 × 10–5 T at its centre. Find the value ofthe current.
34.
An electron makes 3 × 105 revolutions per second in a circle of radius 0.5 angstrom. Find the magnetic field B at the centre of the circle.
35.
A conducting circular loop of radius a is connected to two long, straight wires. The straight wires carry a current i as shown in figure. Find the magnetic field B at the centre of the loop. ,d a f=kT;k dk o`Ùkkdkj pkyd ywi nks yEcs ,oa lh/ks rkjksa ls tqM+k gqvk gSA lh/ks rkjksa ls fp=kkuqlkj i /kkjk izokfgr gks jgh
gSA ywi ds dsUnz ij pqEcdh; {ks=k Kkr dhft;sA
manishkumarphysics.in
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Chapter # 35
Magnetic Field due to A current i
i
36.
Two circular coils of radii 5.0 cm and 10 cm carry equal currents of 2.0 A. The coils have 50 and 100 turns respectively and are placed in such a way that their planes as well as the centre coincide. Find the magnitude of the magnetic field B at the common centre of the currents in the coils are (a) in the same sense (b) in the opposite sense. 5.0 lseh ,oa 10 lseh f=kT;k okyh o`Ùkkdkj dq.Mfy;ksa ls 2.0 A dh leku /kkjk,¡ izokfgr gks jgh gSA buesa Qsjksa dh la[;k,¡ Øe'k% 50 ,oa 100 gSA dq.Mfy;ksa dks bl izdkj j[kk x;k gS fd muds dsUnz rFkk ry nksuksa lEikfrr gSA dq.Mfy;ksa ds mHk;fu"B dsUnz ij pqEcdh; {ks=k dk eku Kkr dhft;sA ;fn dq.Mfyksa esa izokfgr /kkjk,¡ (a) ,d gh fn'kk esa gSA (b) foifjr fn'kk esa gSA Ans. zero 'kwU;
37.
If the outer coil of the previous problem is rotated through 90º about a diameter, what would be the magnitude of the magnetic field B at the centre? ;fn fiNys iz'u esa ckgjh dq.Myh dks blds O;kl ds ifjr% 90º ls ?kqek fn;k tk;sA dsUnz ij pqEcdh; {ks=k B dk ifjek.k
fdruk gksxk\ Ans. 38.
1.8 mT
A circular loop of radius 20 cm carries a current of 10 A. An electron crosses the plane of the loop with a speed of 2.0 × 106 m/s. The direction of motion makes an angle of 30° with the axis of the circle and passes through its centre. Find the magnitude of the magnetic force on the electron at the instant it crosses the plane. 20 lseh f=kT;k okys ,d o`Ùkkdkj ywi ls 10 A /kkjk izokfgr gks jgh gSA ywi ds ry dks ,d bysDVªkuW 2.0 × 106 eh/ls- pky ls ikj djrk gSA xfr dh fn'kk o`Ùk dh v{k ls 30° dks.k cukrh gS rFkk blds dsUnz ls xqtjrh gSA ftl {k.k bysDVªkWu ywi
ds ry dks ikj djrk gSA bl ij pqEcdh; cy dk ifjek.k Kkr dhft;sA 39.
A circular loop of radius R carries a current I. Another circular loop of radius r (<< R ) carries a current i and is placed at the centre o the larger loop. The planes of the two circles are at right angle to each other. find the torque acting on the smaller loop. R f=kT;k okys o`Ùkkdkj ywi ls /kkjk izokfgr gks jgh gSA ,d vU; ywi ftldh f=kt;k r (<< R ) gS rFkk i /kkjk izokfgr gks jgh
gSA cM+s ywi ds dsUnz ij j[kk gqvk gSA nksuksa o`Ùkksa ds ry ijLij yEcor~ gSA NksVs ywi ij yxus okyk cy vk?kw.kZ Kkr dhft;sA 40.
A circular loop of radius r carrying a current i is helf at the centre of another circular loop of radius R(>>r) carrying a current . The plane of the smaller loop makes an angle of 30º with that of the larger loop. If the smaller loop is held fixed in this position by applying a single force at a point on its periphery, what would be the minimum magnitude of this force? r f=kT;k ds o`Ùkkdkj ywi ls i /kkjk izokfgr gks jgh gS] bldks R(>>r) f=kT;k ds ,d vU; o`Ùkkdkj ywi ftlls /kkjk izokfgr gks jgh gS] ds dsUnz ij j[kk x;k gSA NksVs ywi dk ry cM+s ywi ds ry ls 30º dks.k cukrk gSA ;fn NksVs ywi dh ifjf/k ds
fdlh fcUnq ij ,d cy yxkdj ywi dks fLFkj j[kk tk;s rks bl cy dk U;wure ifjek.k fdruk gksuk pkfg;sA Ans. 41.
0 ir 4R
Find the magnetic field B due to a semicircular wire of radius 10.0 cm carrying a current of 5.0 A at its centre of curvature. 10.0 lseh f=kT;k ds v/kZoÙ` kkdkj rkj esa 5.0 A /kkjk izokfgr gks jgh gS] blds oØrk dsUnz ij pqEcdh; {ks=k B dk ifjek.k Kkr
dhft;sA 42.
A piece of wire carryin a current of 6.00 A is bent in the form of a circular arc of radius 10.0 cm, and it subtends an angle of 120° at the centre. Find the magnetic field B due to this piece of wire at the centre. ,d rkj ds VqdM+s ls 6.00 A /kkjk izokfgr gks jgh gS] bldks 10.0 lseh f=kT;k ds o`Ùkkdkj pki ds :i esa eksM+k x;k gS tks dsUnz ij 120° dks.k fufeZr djrk gSA bl VqdM+s ds dkj.k dsUnz ij pqEcdh; {ks=k Kkr dhft;sA
43.
A circular loop of radius r carries a current i. How should a long, straightr wire carrying a current 4i be placed in the plane of the circle so that the magnetic field at the centre becomes zero? r f=kT;k ds o`Ùkkdkj ywi ls i /kkjk izokfgr gks jgh gSA ,d yEcs ,oa lh/ks rkj] ftlls 4i /kkjk izokfgr gks jgh gS] dks o`Ùk ds manishkumarphysics.in
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Chapter # 35
Magnetic Field due to A current
ry esa fdl izdkj j[kk tk;s fd dsUnz ij pqEcdh; {ks=k 'kwU; gks tk;s\ Ans. at a distance of 4r/ from the centre in such a way that the direction of the current in it is opposite to that in the nearest part of the circular wire 44.
A circular coil of 200 turns has a radius of 10 cm and carries a current of 2.0 A. (a) Find the magnitude of the magnetic field B at the centre of the coil. (b) At what distance from the centre along the axis of the coil will the field B drop to half its value at the centre ? ( 3 4 = 1.5874 ....) 200 Qsjksa ,oa 10 lseh f=kT;k okyh o`Ùkkdkj dq.Myh ls 2.0 A /kkjk izokfgr gks jgh gSA (a) dq.Myh ds dsUnz ij pqEcdh; {ks=k B Kkr dhft;sA (b) dq.Myh dh v{k ij dsUnz ls fdruh nwjh ij fLFkr fcUnq ij pqEcdh; {ks=k dk eku B, dsUnz ij pqEcdh;
{ks=k ds eku dk vk/kk jg tk;sxk\ ( 3 4 = 1.5874 ....) 45.
A circular loop of radius 4.0 cm is placed in a horizontal plane and carries an electric current of 5.0 A in the clockwise direction as seen from above. find the magnetic field (a) at a point 3.0 cm above the centre of the loop (b) at a point 3.0 cm below the centre of the loop. {kSfrt ry esa j[ks gq, 4.0 lseh f=kT;k ds ,d o`Ùkkdkj ywi ls 5.0 A /kkjk izokfgr gks jgh gSA Åij ls ns[kus ij /kkjk dh fn'kk nf{k.kkorZ gSA pqEcdh; {ks=k Kkr dhft;sA (a) ywi ds dsUnz ls 3.0 lseh Åij fLFkr fcUnq ij (b) ywi ds dsUnz ls 3.0 lseh
uhps fLFkr fcUnq ijA
Ans. : 4.0 × 10–5 T, downwards in both the cases 46.
A charge of 3.14 × 10–6 C is distributed uniformly over a circular ring of radius 20.0 cm. The ring rotates about its axis with an angular velocity of 60.0 rad/s. Find the ratio of the electric field to magnetic field at a point on the axis at a distance of 5.00 cm from the centre. 20.0 lseh f=kT;k dh ,d o`Ùkkdkj oy; ij 3.14 × 10–6 dwykWe vkos'k ,d leku :i ls forfjr gSA oy; dks bldh v{k ds ifjr% 60.0 jsfM;u@ls- dks.kh; pky ls ?kqek;k tkrk gSA dsUnz ls 5.00 lseh nwj fLFkr v{kh; fcUnq ij fo|qr {ks=k rFkk
pqEcdh; {ks=k dk vuqikr Kkr dhft;sA Ans. 47.
A thin but long, hollow, cylindrical tube of radius r carries a current i along its length. Find the magnitude of the magnetic field at a distance r/2 from the surface (a) inside the tube (b) outside the tube. ,d iryh fdUrq yEch] [kks[kyh csyukdkj ufydk dh f=kT;k r gS] blls yEckbZ ds vuqfn'k i /kkjk izokfgr gks jgh gSA lrg ls r/2 nwjh ij pqEcdh; {ks=k Kkr dhft;sA (a) ufydk ds vUnj (b) ufydk ds ckgj 0i 2b A long, cylindrical tube of inner and outer radii a and b carries a current i distributed uniformly over its crosssection. Find the magnitude of the magnetic field at a point (a) Just inside the tube (b) just outside the tube. ,d yEch csyukdkj ufydk dh vkarfjd ,oa cká f=kT;k,¡ Øe'k% a rFkk b gSA ufydk ls izokfgr i /kkjk blds vuqiLz Fk dkV esa ,d leku :i ls forfjr gSA (a) ufydk ds rqjUr vUnj (b) ufydk ds ckgj
Ans. 48.
1.88 × 10–15 m/s
(a) zero (b)
49.
A long, cylindrical wire of radius b carries a current i distributed uniformly over its cross-section. Find the magnitude of the magnetic field at a point inside the wire at a distance a from the axis. b f=kT;k okys ,d yEcs csyukdkj rkj ls izokfgr gksus okyh /kkjk i ] blds vuqiLz Fk&dkV esa ,d leku :i ls forfjr gSA rkj ds vUnj v{k ls a nwjh ij fLFkr fcUnq ij pqEcdh; {ks=k dk ifjek.k Kkr dhft;sA
50.
A solid wire of radius 10 cm carries a current of 5.0 A distributed uniformly over its cross-section. Find the magnetic field B at a point at a distance (a) 2 cm (b) 10 cm and (c) 20 cm away from the axis. Sketch a graph of B versus x for 0 < x < 20 cm. 10 lseh f=kT;k okys Bksl rkj ls izokfgr 5.0 A /kkjk blds vuqiLz Fk dkV esa ,d leku :i ls forfjr gSA v{k ls (a) 2 lseh (b) 10 lseh rFkk (c) 20 lseh nwj fLFkr fcUnqvksa ij pqEcdh; {ks=k dk ifjek.k Kkr dhft;sA 0 < x < 20 lseh ds fy;s x ds lkis{k B dk ys[kkfp=k Hkh cukb;s A
51.
Some time we show an idealised magnetic field which is uniform in a given region and falls to zero abruptly. One such field is represented in figure. Using Ampere's law over the path PQRS, show that such a field is not possible.
dHkh&dHkh ge ,d ,slk vkn'kZ pqEcdh; {ks=k iznf'kZr djrs gSa tks fns;s x;s {ks=k esa le:i gksrk gS rFkk ,dne ls 'kwU; gks tkrk gSA ,slk gh ,d {ks=k fp=k esa iznf'kZr fd;k x;k gSA iFk PQRS ds fy;s ,Eih;j ds fu;e dk mi;ksx djds] O;Dr dhft;s fd ,slk {ks=k laHko ugha gSA manishkumarphysics.in
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Chapter # 35 52.
Magnetic Field due to A current
Two large metal sheets carry surface currents as shown in figure. The current through a strip of width dl is Kdl where K is a constant. Find the magnetic field at the point P, Q and R. /kkrq dh nks fo'kky ifV~Vdkvksa esa fp=kkuqlkj lrg /kkjk,¡ izokfgr gks jgh gSA d pkSM+kbZ dh ,d iV~Vh ls Kd /kkjk izokfgr gks jgh gSA tgk¡ K fu;rkad gSA fcUnqvksa P, Q rFkk R ij pqEcdh; {ks=k Kkr dhft;sA P
Q
× × × × × × × × R
Ans. 53.
0, 0 K towards right in the figure, 0
Consider the situatio of the previous problem. A particle having charge q and mass m is projected from the point Q in a direction going into the plane of the diagram. It is found to describe a circle of radius r between the two plates. Find the speed of the charged particle. fiNys iz'u esa of.kZr ifjfLFkfr ij fopkj dhft;sA m nzO;eku rFkk q vkos'k okyk ,d d.k fcUnq Q ls fp=k ds ry ds vUnj tkus okyh fn'kk esa iz{ksfir fd;k tkrk gSA d.k nksuksa ifV~Vdkvksa ds e/; r f=kT;k ds o`Ùkkdkj iFk ij xfr djrk gSA vkosf'kr
d.k dh pky Kkr dhft;sA 54.
The magnetic field B inside a long solenoid, carrying a current of 5.00 A, is 3.14 × 10–2 T. Find the number of turns pwr unit length of the solenoid. ,d yEch ifjufydk esa 5.00 A /kkjk izokfgr gks jgh gSA blds vUnj pqEcdh; {ks=k B dk ifjek.k 3.14 × 10–2 VsLyk gSA
ifjufydk dh ,dkad yEckbZ esa Qsjksa dh la[;k Kkr dhft;sA Ans.
5000 turns/m
55.
A long solenoid is fabricated by closely winding a wire of radius 0.5 mm over a cylindrical nonmangetic frame so that the successive turns nearly touch each other. What would be the magnetic field B at the centre of the solenoid if it carries a current of 5 A? ,d yEch vpqEcdh; Ýse ij 0.5 feeh f=kT;k dk rkj ikl&ikl yisV dj ,d yEch ifjufydk cukbZ tkrh gS] rkj dks bl izdkj yisVk tkrk gS fd Øekxr Qsjs ,d&nwljs dks Li'kZ djrs gSaA ;fn blls 5 A /kkjk izokfgr gks jgh gks rks ifjufydk ds dsUnz ij pqEcdh; {ks=k B dk eku fdruk gksxk\ Ans. 2 × 10–3 T
56
A copper wire having resistance 0.01 ohm in each metre is used to wind a 400 turn solenoid of radius 1.0 cm and length 20 cm. Find the emf of a battery which when connected across the solenoid will cause a magnetic field of 1.0 × 10–2 T near the centre of the solenoid. ,d rkacs ds rkj dh bdkbZ yEckbZ dkizfrjks/k 0.01 vkse gS] bldk mi;ksx djds 1.0 lseh f=kT;k rFkk 20 lseh yEckbZ dh ifjufydk ds 400 Qsjs yisVs tkrs gSA ml cSVjh dk fo-ok-cy Kkr dhft;sA ftldks bl ifjufydk ls tksMu+ s ij] ifjufydk ds dsUnz ij 1.0 × 10–2 VsLyk dk pqEcdh; {ks=k mRiUu gksA Ans. 2 × 10–3 T
57.
A tightly wound solenoid of radius a and length l has n turens per unit length. It carries an electric current i. Consider a length dx of the solenoid at a distance x from one end. This contians n dx turns and may be approximated as a circular current in dx. (a) Write the magnetic field at the centre of the solenoid due to this circular current. Integrate this expression under proper limits to find the magnetic field at the centre of the solenoid. (b) Varify that if l >> a, the field tends to B = 0 ni and if a >> 1, the field tends to B =
nil . Interpret 2a
these results.
,d dl dj yisVh gqbZ ifjufydk dh f=kT;k a rFkk yEckbZ gSA bldh izfr bdkbZ yEckbZ esa Qsjksa dh la[;k n gSA blls i /kkjk izokfgr gks jgh gSA ifjufydk ds ,d fljs ls x nwjh ij dx yEckbZ ij fopkj dhft;sA blesa n dx Qsjs gSa rFkk bldks indx eku dh o`Ùkkdkj /kkjk ds rqY; ekuk tk ldrk gSA (a) bl o`Ùkkdkj /kkjk ds dkj.k ifjufydk ds dsUnz ij pqEcdh; {ks=k Kkr djus ds fy;s bl O;atd dk mfpr lhekvksa esa lekdyu dhft;sA (b) tk¡p dhft;s fd ;fn >> a, rks {ks=k B = 0 ni rFkk Ans.
(a)
;fn a >> , rks {ks=k B =
nil 2a
ds rqY; gks tkrk gSA vius mÙkj dh O;k[;k dhft;sA
ni 2a 1
2
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Chapter # 35 Magnetic Field due to A current 58. A tightly wound, long solenoid carries a current of 2.00 A. An electron is found to execute a uniform circular motion inside the solenoid with a frequency of 1.00 × 108 rev/s. Find the number of turns per metre in the solenoid. ,d dldj yisVh gqbZ yEch ifjufydk esa 2.00 A /kkjk izokfgr gks jgh gSA ifjufydk esa ,d bysDVªkWu 1.00 × 108
[email protected] dh vko`fÙk ls ,d leku o`Ùkh; xfr djrk gqvk ik;k tkrk gSA ifjufydk esa izfr ehVj Qsjksa dh la[;k Kkr dhft;sA Ans. : 1240
59.
A tightly-would, long solenoid has n turns per unit length, a radius r and carries a current i. A particle having charge q and mass m is projected from a point on the axis in a direction perpendicular to the axis. What can be the maximum speed for which the particle does not strike the solenoid? ,d dldj fyiVh gqbZ ifjufydk dh bdkbZ yEckbZ esa n Qsjs gS] bldh f=kT;k r gS rFkk blesa i /kkjk izokfgr gks jgh gSA bldh v{k ij fLFkr ,d fcUnq ls m nzO;eku ,oa q vkos'k okyk ,d d.k v{k ds yEcor~ fn'kk esa iz{ksfir fd;k tkrk gSA vf/kdre
pky dk eku fdruk gks ldrk gS fd d.k ifjufydk ls u Vdjk,\ Ans. 60.
0 ni 2m
A tightly-wound, long solenoid is kept with its axis parallel to a large metal sheet carrying a surface current. The surface current through a width d of the sheet is Kd and the number of turns per unit length of the solenoid. (b) If the solenoid is rotated to make its axis perpendicular to the metal sheet, what would be the magnitude of the magnetic field near its centre?
,d fo'kky /kkrq dh ifV~Vdk esa i`"Bh; /kkjk izokfgr gks jgh gS] blds lehi ,d dldj yisVh gqbZ yEch ifjufydk bl izdkj j[kh gqbZ gS fd bldh v{k ifV~Vdk ds lekukUrj gSA ifV~Vdk dh d pkSM+kbZ ls i`"B /kkjk Kd gS rFkk ifjufydk dh ,dkad yEckbZ esa Qsjksa dh la[;k n gSA ifjufydk ds dsUnz ds lehi pqEcdh; {ks=k 'kwU; gSA (a) ifjufydk esa izokfgr /kkjk Kkr dhft;sA (b) ;fn ifjufydk dks bl izdkj ?kqek;k tkrk gS fd bldh v{k /kkrq dh ifV~Vdk ds yEcor~ gks tk;s] rks blds dsUnz ds lehi pqEcdh; {ks=k dk ifjek.k fdruk gksxk\ Ans. 61.
(a)
0 K K (b) 2 2n
A capacitor of capacitance 100 F is connected to a battery of 20 volts for a long time and then disconnected from it. It is now connected across the capacitor drops to 90% of its malximum value in 2.0 seconds. Estimate the average magnetic field produced at the centre of the solenoid during this period. 100 F /kkfjrk dk la/kkfj=k] 20 oksYV dh cSVjh ls yEcs le;rd la;ksftr j[kus ds i'pkr~ blls vyx dj fy;k tkrk gSA bldks vc rd yEch ifjufydk ls tksM+k tkrk gSA ftlls 4000 Qsjs izfr ehVj gSA ;g Kkr gksrk gS fd 2.0 lsd.M i'pkr~ la/kkfj=k ds fljksa ij foHkokUrj izkjfEHkd eku dk 90% jg tkrk gSA bl le;kUrj esa ifjufydk ds dsUnz ij mRiUu vkSlr
pqEcdh; {ks=k dk vuqeku yxkb;sA Ans.
16 × 10–8 T.
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