Chapter # 34
34.1
Magnetic Field
SOLVED EXAMPLES
A proton is projected with a speed of 3 × 106 m/s horizontally from east to west. A uniform magnetic field B of strength 2.0 × 10–3 T exists in the vertically upward direction. (a) Find the force on the proton just after it is projected. (b) What is the acceleration produced?
Sol.
B F
N
v
E
W S
(a) The situation is shown in figure. The force is perpendicular to B hence it is in the horizontal plane through the proton. In this plane, it is perpendicular to the velocity v . Thus, it is along the north-south line. The rule for vector product shows that × B is towards north. As the charge on the proton is positive, the force F qv B is also towards north. The magnitude of the force is F = qvB sin = (1.6 × 10–19 C) (3.0 × 106 m/s) (2.0 × 10–3 T) = 9.6 × 10–16 N. (b) The acceleration of the proton is a=
9.6 10 16 N F = m 1.67 10 27 kg
= 5.8 × 1011 m/s2. 34.2
Sol.
A particle having a charge of 100 C and a mass of 10 mg is projected in a uniform magnetic field of 25 mT with a speed of 10 m/s. If the velocity is perpendicular to the magnetic field, how long will it take for the particle to come back to its original position for the first time after being projected. The particle moves along a circle and returns to its original position after completing the circle, that is after one time period. The time period is T=
= 34.3
2m qB 2 (10 10 9 kg)
(100 10 6 C) (25 10 3 T )
= 25 a.
Figure shows a triangular loop PQR carrying a current i. The triangle is equilateral with edge-length l. A uniform magnetic field B exists in a direction parallel to PQ. Find the forces acting on the three wire PQ, QR and RP separately.
R
i
P Sol.
B
i
i
Q
The force on the wire PQ is F1 iPQ B 0 as the field B is parallel to PQ . The force on QR is manishkumarphysics.in
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Chapter # 34
Magnetic Field F2 i PQ B 0
or,
F2 = iB sin 120º
3 iB . 2 From the rule of vector product, this force is perpendicular to the plane of the diagram and is going into it. The force on RP is F3 i RP B =
3 iB. 2 From the rule of vector product, this force is perpendicular to the plane of the diagram end is coming out of it. or,
34.4 Sol.
F3 = i B sin 120º =
A current of 10.0 nA is established in a circular loop of radius 5.0 cm. Find the magnetic dipole moment of the current loop. The magnetic dipole moment is iA . Thus, = ir3 = (10 × 10–9 A) (3.14) × (5 ×10–2 m)2. = 7.85 × 10–11 A -m2 .
QUESTIONS 1.
2. 3. 4. 5. 6.
7. 8. 9. 10.
FOR
SHORT
ANSWER
Suppose a charged particle moves with a velocity v near a wire carrying an electric current. A magnetic force, therefore, acts on it. If the same particle is seen from a frame moving with velocity v in the same direction, the charge will be found at rest. Will the magnetic force become zero in this frame ? Will the magnetic field become zero in this frame ? Can a charged particle be accelerated by a magnetic field ? Can its speed be increased ? Will a current loop placed in a magnetic field always experiment a zero force ? The free electrons in a conducting wire are in constant thermal motion. If such a wire, carrying no current, is placed in a magnetic field, is there a magnetic force on each free electron ? On the wire ? Assume that the magnetic field is uniform in a cubical region and is zero outside. Can you project a charged particle from outside into the field so that the particle describes a complete circle in the field ? An electron beam projected along the positive X–axis deflects along the positive Y–axis. If this deflection is caused by a magnetic field, what is the direction of the field ? Can we conclude that the field is parallel to the Z–axis ? Is it possible for a current loop to stay without rotating in a uniform magnetic field > If yes, what should be the orientation of the loop ? The net charge in a current–carrying wire is zero. Then, why does a magnetic field exert a force on it ? The torque on a current loop is zero if the angle between the positive normal and the magnetic field is either = 0 or = 180º. In which of the two orientations, the equilibrium is stable ? Verify that the units weber and volt–second are the same.
Objective - I 1.
A positively charged particle projected towards east os deflected towards north by a magnetic field. The field may be (A) towards west (B) towards south (C) upward (D*) downward iwoZ dh vksj iz{ksfir ,d /kukosf'kr d.k pqEcdh; {ks=k ds dkj.k mÙkj dh vksj fo{ksfir gks tkrk gSA {ks=k gks ldrk gS (A) if'pe dh vksj (B) nf{k.k dh vksj (C) Åij dh vksj (D*) uhps dh vksj
2.
A charged particle is whirled in a horizontal circle on a frictionless table by attaching it to a string fixed at one point. If a magnetic field is switched on in the vertical direction, the tension in the string. (A) will increase (B) will decrease (C) will remain the same (D*) may increase or decrease
,d vkosf'kr d.k ls Mksjh cka/k dj bldks ,d ?k"kZ.k jfgr Vscy ij {ksfrt o`Ùk esa ?kqek;k tkrk gS] Mksjh dk ,d fljk ,d fcUnq ij fLFkj jgrk gSA ;fn m/okZ/kj fn'kk esa pqEcdh; {ks=k vkjksfir fd;k tk;s] Mksjh esa ruko -
RESONANCE
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Chapter # 34 (A) c<+sxk (D*) vf/kd 3.
Magnetic Field (B) de gksxk ;k de gks ldrk gSA
(C) leku
jgsxk
Which of the following particles will describe the smallest circle when projected with the same velocity perpendicular to a magnetic field ? (A) electron (B) proton (C) He+ (D*) Li+
pqEcdh; {ks=k ds yEcor~ leku osx ls iz{ksfir djus ij fuEu esa ls dkSulk d.k vf/kdre pqEcdh; cy ¼ifjek.k es½a vuqHko djsxk (A) bysDVªkWu (B) izksVkWu (C) He+ (D*) Li+ 4.
Which of the following particles will have minimum frequency of revolution when projected with the same velocity perpendicular to a magnetic field ? (A*) electron (B) proton (C) He+ (D) Li+ pqEcdh; {ks=k ds yEcor~ leku osx ls iz{ksfir djus ij dkSulk d.k lcls NksVs o`Ùk ij xfr djsxk (A*) bysDVªkWu (B) izksVkWu (C) He+ (D) Li+
5.
Which of the following particles will have minimum frequency of revolution when projected with the same velocity perpendicular to a magnetic field ? (A) electron (B) proton (C) He+ (D*) Li+ pqEcdh; {ks=k ds yEcor~ leku osx ls iz{ksfir djus ij fuEu esa ls dkSuls d.k dh ifjØe.k vko`fÙk U;wure gksxh (A) bysDVªkWu (B) izksVkWu (C) He+ (D*) Li+
6.
A circular loop of area 1 cm2, carrying a current of 10 A, is placed in a magnetic field of 0.1 T perpendicular to the plane of the loop. The torque on the loop due to the magnetic field is 1 lseh2 {ks=kQy okyk o`Ùkkdkj ywi] ftlesa 10 A /kkjk izokfgr gks jgh gS] 0.1 VsLyk rhozrk ds pqEcdh; {ks=k esa j[kk gqvk gS] pqEcdh; {ks=k ds dkj.k ywi ij cy vk?kw.kZ gS (A*) zero 'kwU; (B) 10-4 N-m (C) 102 N-m (D) 1 N-m
7.
A beam consisting of protons and electrons moving at the same speed goes through a thin region in which there is a magnetic field perpandicular to the beam. The protons and the electrons (A) will go undeviated (B) will be deviated by the same angle will not separate (C*) will be deviated by different angles and hence separate (D) will be deviated by the same angle but will separate.
leku pky ls xfr'khy bysDVªkWuksa rFkk izkVs kWuksa dk ,d iqt a ] ,d ladh.kZ ,sls {ks=k ls xqtjrk gS] tgk¡ pqEcdh; {ks=k dh fn'kk iqat ds yEcor~ gSA izkVs kWu rFkk bysDVªkWu (A) fcuk fo{ksfir gq, pys tk;sxAsa (B) leku dks.k ls fo{ksfir gksaxs rFkk i`Fkd ugha gksx a As (C*) fHkUu&fHkUu dks.kksa ls fopfyr gksx]as vr% i`Fkd gks tk;sxAas (D) leku dks.k ls fo{ksfir gksx]as fdUrq i`Fkd gks tk;sxAas 8.
A charged particle moves in a uniform magnetic field. The velocity of the particle at some instant makes an acute angle with the magnetic field. The path of the particle will be. (A) a straight line (B) a circle (C*) a helix with uniform pitch (D) a helix with nonuniform pitch.
,d vkosf'kr d.k le:i pqEcdh; {ks=k esa xfr djrk gSA fdlh {k.k ij d.k ds osx dh fn'kk] pqEcdh; {ks=k ls U;wu dks.k cukrh gSA d.k dk iFk gksxk (A) ljy js[kk (B) o`Ùk (C*) ,d leku pwM+h varjky okyk gsfydl (D) vleku pwM+h varjky okyk gsfydl 9.
A particle moves in a region having a uniform magnetic field and a parallel, uniform electric field. At some instant, the velocity of the particle is perpendiculars to the field direction. The path of the particle will be (A) a straight line (B) a circle (C) a helix with uniform pitch (D*) a helix with nonuniform pitch.
,d d.k ,sls {ks=k esa xfr djrk gS] tgk¡ le:i pqEcdh; {ks=k gS rFkk lekukarj fo|qr {ks=k gSA fdlh {k.k ij d.k ds osx dh fn'kk] {ks=k dh fn'kk ds yEcor~ gSA d.k dk iFk gksxk (A) ljy js[kk (B) o`Ùk (C) ,d leku pwM+h varjky okyk gsfydl (D*) vleku pwM+h varjky okyk gsfydl manishkumarphysics.in
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Chapter # 34 Magnetic Field 10. An electric current i enters and leaves a uniform circular wire of radius a through diametrically opposite points. A charged particle q moving along the axis of the circular wire passes through its centre at speed . The magnetic force acting on the particle when it passes through the centre has a magnitude ,d le:i o`Ùkkdkj rkj dh f=kT;k a gS] blesa O;klr% vfHkeq[k fcUnqvksa ls i /kkjk izo's k djrh gS rFkk ckgj fudyrh gSA ,d vkosf'kr d.k q o`Ùkkdkj rkj dh v{k ds vuqfn'k pky ls xfr djrk gqvk blds dsUnz ls xqtjrk gSA tc d.k dsUnz ls xqtjrk gS rks bl ij yxus okys pqEcdh; cy dk ifjek.k gksxk (A) q
0i 2a
(B) q
0 i 2a
(C) q
0i a
(D*) zero
Objective - II 1.
If a charged particle at rest experieces no electromagnetic force, (A*) the electric field must be zero (B) the magnetic field must be zero (C) the electric field may or may not be zero (D*) the magnetic field may or may not be zero ;fn fojkekoLFkk esa fLFkr d.k dksbZ fo|qr&pqEcdh; cy vuqHko ugha djrk gS (A*) fo|qr {ks=k 'kwU; gksxkA (B) pqEcdh; {ks=k 'kwU; gksxkA (C) fo|qr {ks=k 'kwU; gks Hkh ldrk gS vkSj ugha HkhA (D*) pqEcdh; {ks=k 'kwU; gks Hkh ldrk gS vkSj ugha HkhA
2.
If a charged particle kept at list expreiences an electromagnetic force, (A*) the electric field must not be zero (B) the magnetic field must not be zero (C) the electric field may or may not be zero (D*) the magnetic field may no may not be zero ;fn fojkekoLFkk esa fLFkr ,d d.k fo|qr&pqEcdh; cy vuqHko djrk gS (A*) fo|qr {ks=k 'kwU; ughs gksxk (B) pqEcdh; {ks=k 'kwU; ughs gksxkA (C) fo|qr {ks=k 'kwU; gks Hkh ldrk gS vkSj ugha HkhA (D*) pqEcdh; {ks=k 'kwU; gks Hkh ldrk gS vkSj ugha HkhA
3.
If a charged particle projected in a gravity-free room deflects, (A) there must be an electric field (B) there must be a magnetic field (C*) both field cannot be zero (D*) both fields can be nonzero ,d xq:Roghu dejs esa iz{ksfir fd;k x;k vkosf'kr d.k fo{ksfir gksrk gS] rks (A) ogk¡ ij fo|qr {ks=k gksxkA (B) ogk¡ ij pqEcdh; {ks=k gksxkA (C*) nksuksa {ks=k 'kwU; ugha gks ldrs gSAa (D*) nksuksa {ks=k v'kwU; gks ldrs gSAa
4.
A charged particle moves in a gravity-free space without change in veloicty. Which of the following is/are possible ? ,d vkosf'kr d.k xq:Ro&eqDr {ks=k esa osx esa ifjorZu ds fcuk xfr djrk gSA fuEu esa ls dkSulk&dkSuls lEHko gS (A*) E = 0, B = 0 (B*) E = 0, B 0 (C) E 0, B = 0 (D*) E 0, B 0
5.
A charged particle moves alonh a circle under the action of possible constant electric and magnetic fields. Which of the following are possible ? ,d vkosf'kr d.k laHko ,oa fu;r fo|qr rFkk pqEcdh; {ks=kksa esa o`Ùkkdkj iFk ij xfr djrk gSA fuEu esa ls dkSulk laHko gS(A) E = 0, B = 0 (B*) E = 0, B 0 (C) E 0, B = 0 (D) E 0, B 0
6.
A charged particle goes undelflected in a region containing electric and magnetic field. It is possible that
(B*) E is not parallel to B E B E (C) B but E is not parallel to B (D) E B but is not parallel to E fdlh {ks=k esa tgk¡ fo|qr ,oa pqEcdh; {ks=k mifLFkr gSA ,d vkosf'kr d.k fo{ksfir gq, fcuk xfr djrk gSA ;g laHko gS fd (A*) E B E (B*) E B ds lekukarj ugha gSA (C) B fdUrq E ] B ds lekukarj ugha gSA (D) E B fdUrq , E ds lekukarj ugha gSA (A*)
7.
If a charged particle goes unacceleration in a region containing electric and magnetic fields, (A*) E must be perpeandicular to B (B*) must be perpendicular to E (C) must be perpendicular to B (D) E must be equla to B. ;fn ,d vkosf'kr d.k ,sls {ks=k esa Roj.kghu xfr djrk gS] tgk¡ fo|qr ,oa pqEcdh; {ks=k gS (A*) E , B ds yEcor~ gksxkA (B*) , E ds yEcor~ gksxkA (C) , B ds yEcor~ gksxkA (D) E, B ds cjkcj gksxkA manishkumarphysics.in
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Chapter # 34 Magnetic Field 8. Two ions have equla masses but one is singly-ionized and other is douly-ionized. The are project from the same place in a uniform magnetic field with the same veloicty perpendicular to the field. (A) Both ions will go along circles of equal radii. (B*) The circle described by the single-ionized charge will have a radius double that of the other circle (C) The two circles do not touch each other (D*) The two circles touch each other
nks vk;uksa ds nzO;eku ,d leku gS] fdUrq ,d ,d/kk vk;fur rFkk nwljk f}&vk;fur gSA budks ,d gh LFkku ls ,d leku pqEcdh; {ks=k esa ,d leku osx ls pqEcdh; {ks=k ds yEcor~ iz{ksfir fd;k tkrk gS (A) nksuksa vk;u leku f=kT;kvksa ds o`Ùkkdkj iFkksa ij xfr djsxaAs (B*) ,d/kk vk;fur vk;u }kjk cuk;s x;s o`Ùkkdkj (C) nksuksa o`Ùk ,d nwljs Li'kZ ugha djsxsaA (D*) nksuksa o`Ùk ,d nwljs dks Li'kZ djsxsaA 9.
10.
An electron is moving along the positive X-axis. You want to apply a magnetic field for a short time so that the electron may reverse its direction and move parallel to the negative X-axis. This can be done by applying the magentic field along. (A*) Y-axis (B*) Z-axis (C) Y-axis only (D) Z-axis only ,d bysDVªkuW /kukRed X-v{k ds vuqfn'k xfr dj jgk gSA vki vYi le; ds fy;s ,dpqEcdh; {ks=k yxkuk pkgrs gS]a rkfd bysDVªkuW viuh fn'kk foijhr djds _.kkRed X-v{k dh vksj xfr dj ldsA ;g fd;k tk ldrk gS ;fn pqEcd; {ks=k fuEu ds vuqfn'k yxk;k tk;s (A*) Y-v{k (B*) Z-v{k (C) dsoy Y-v{k (D) dsoy Z-v{k Let E and B denote electric and magnetic fields in a frames S and E ’ and B ’ in another frame S’ moving
with respect to S at a velocity . Two of the following equations are wrong. Identify them. ekuk fd E rFkk B ,d funs'Z k ra=k S esa fo|qr rFkk pqEcdh; {ks=kksa dks O;Dr djrs gSa rFkk S ds lkis{k nwljs funsZ'k ra=k S’ esa E rFkk B A fuEu esa ls nks lehdj.ksa xyr gSA budks fpfUgr dhft;s (A) By1 + By +
E 2 c2
(B*) Ey1 + Ey +
B2 c2
(C*) By1 + By + E2
osx ls xfr'khy
(D) By1 + By + B2
WORKED OUT EXAMPLES
1.
Sol.
A charge of 2.0 C moves with a speed of 2.0 × 106 m/s along the positive X-axis. A magnetic field B of strength (0.20 j + 0.40 k )T exists in space. Find the magnetic force acting on the charge. The force on the charge = q v × B
= (2.0 × 10–6) (2.0 × 10 6 m/s i ) × (0.20 j + 0.40 k )T = 4.0(0.20 i × j + 0.40 i × k )N = (0.8 k – 1.6 j ) N.
2.
A wire is bent in the form of an equilateral triangle PQR of side 10 cm and carries a current of 5.0 A. It is placed in a magnetic field B of magnitude 2.0 T directed perpendicularly to the plane of the loop. Find the forces on the three sides of the triangle.
Sol.
R
B
F3 F2
F1 P
i
Q
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Chapter # 34 Magnetic Field Suppose the field and the current have directions as shown in figure. The force on PQ is F1 i B or, F1 = 5.0 A × 10 cm × 2.0 T = 1.0 N The rule of vector product shows that the force F1 is perpendicular to PQ and is directed towards the inside of the triangle. The forces F2 and F3 on QR and RP can also be obtained similarly. Both the forces are 1.0 N directed perpendicularly to the respective sides and towards the inside of the triangle. The three forces F1 , F2 and F3 will have zero resultant, so that there is no net magnetic force on the triangle. This result can be generalised. Any closed current loop, placed in a hogogeneous magnetic field, does not experience a net magnetic force. 3.
Figure shows two long metal rails placed horizontally and parallel to each other at a separation l. A uniform magnetic field B exists in the vertically downward direction. A wire of mass m can slide on the rails. The rails are connected to a constant current source which drives a current i in the circuit. The friction coefficient between the rails and the wire is . (a) What soluble the minimum value of which can prevent the wire from sliding on the rails? (b) Describe the motion of the wire if the value of is half the value found in the previous part
i
Sol.
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
(a) The force on the wire due to the magnetic field is F i B or, F = iB It acts towards right tin the given figure. If the wire does not slide on the rails, the force of friction by the rails should be equal to F. If 0 be the minimum coefficient of friction which can prevent sliding, this force is also equal to 0 mg. Thus, 0 mg = iB 0 =
or,
iB mg
(b) If the friction coefficient is =
iB 0 = 2 mg , the wire will slide towards right. The frictional force by the rails 2
is f = mg =
iB towards left. 2
iB iB iB = towards right. The acceleration will be a = . The wire will slide 2m 2 2 towards right with this acceleration. The resultant force is iB –
4.
A proton, a deuteron and an alpha particle moving with equal kinetic enter perpendicularly into a magnetic field. If rp, rd and ra are the respective radii of the circular paths, find the ratio rp/ rd and rp/rn.
Sol.
We have r = where K =
mv = qB
2mK qB
1 mv2 = kinetic energy.. 2
Thus,
ip =
and
ra =
2mpK qpB 2m aK q aB
,
rd =
2m dK q dB
. manishkumarphysics.in
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Chapter # 34
Magnetic Field
rp We get
rd rp
and
5.
Sol.
qd qp
md
qa = q p
ra
qp = q p
mp
mp ma
mp
1 =
ma
2qp = q p
2
mp
4 mp = 1.
Singly charged magnesium (A = 24) ions are accelerated to kinetic energy 2 keV and are projected perpendicularly into a magnetic field B of magnitude 0.6 T. (a) Find the radius of the circle formed by the ions. (b) If there are also singly charged ions of the isotope magnesium-26, what would be the radius for these particles? The radius is given by r=
mv = qB
2 mK qB
For 24Mg ions, m = 24 × mp approximately and q = 1.6 × 10–10 C. Putting the values, r=
2 24 1.67 10 27 kg 2000 1.6 10 18 J 1.6 10 19 C 0.6 T
= 0.053 m = 5.3 cm. For 28Mg, the radius r’ will be given by r’ =
or, 6.
r’ = r
2m' K qB
m' = 5.3 cm m
26 = 5.5 cm. 24
A particle having a charge 20 C and mass 20 g moves along a circle of radius 5.0 cm under the action of a magnetic field B = 1.0 T. When the particle is at a point P, a uniform electric field is switched on and it is found that the particle continues on the tangent through P with a uniform velocity. Find the electric field.
Sol.
×
×
×
×
×
×
×
×
×
×
×
×
×
×v
×
×
×
×
×
×
×
×
×
P
×
E
×
When the particle moves along a circle in the magnetic field B, the magnetic force is radially inward. If an electric field of proper magnitude is switched on which is directed radially outwards, the particle may experience no force. It will then move along a straight line with uniform velocity. This will be the case when qE = qvB or, E = vB, The radius of the circle in a magnetic field is given by
or,
r=
mv qB
v=
rqB m
(5.0 10 2 m) (20 10 6 C)(1.0 T ) =
20 10 9 kg
= 50 m/s.
The required electric field is E = vB = (50 m/s) (10 × 10–3 T) This field will be in a direction which is radially outward at P.
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Chapter # 34 Magnetic Field 7. A particle of mass m = 1.6 × 10–27 kg and charge q = 1.6 × 10–19 C moves at a speed of 1.0 × 10–7 m/s. It enters a region of uniform magnetic field at a point E, as shown in figure. The field has a strength of 1.0 T. (a) The magnetic field is directed into the plane of the paper. The particle leaves the region of the field at the point F. Find the distance EF and the angle . (b) If the field is coming out of the paper, find the time spent by the particle in the region of the magnetic field after entering it at E.
F
O
E A Sol.
45º
C
×
×
×
×
× ×
× ×
×
×
×
×
× ×
× ×
(a) As the particle enters the magnetic field, it will travel in a circular path. The centre will be on the line
mv . The direction of the force av B shows that the qB centre will be outside the field as shown in figure. As AEO = 90º (as AE is tangent and OE is radius) and AEC = 45º, we have ÐOEF = 45º. As OE = OF (they are radii of the circular arc), OFE = 45º. Also, OF is perpendicular to the velocity of the particle at F, so that = 45º. From triangle OEF, EF = 2.OE cos OEF perpendicular to its velocity and the radius r will be
= 2.
1 mv . qB 2
2 (1.6 10 27 kg) (107 m / s) =
(1.6 10 19 C) 1.0 T –1 2 × 10 m = 14 cm.
=
E
(b)
×
×
×
×
×
E×
×
×
×
×
×
×
×
×
×
×
×
×O
×
×
×
×
×
×
×
F×
×
×
×
×
×
×
×
×
×
If the magnetic field is coming out of the paper, the direction of the force qv × B shows that the centre O will be inside the field region as shown in figure. Again AEO = 90º, giving OEF = OFE = 45º. Thus, the angle EOF = 90º. The particle describes three fourths of the complete circle inside the field. As the speed v is uniform, the time spent in the magnetic field will be
3 2r 3mv 3m 4 v 2vqB 2qB 3 3.14 1.6 10 27 kg =
8.
Sol.
2 1.6 10 19 C 1.0 T
= 4.7 × 10–8 s,
A beam of protons with a velocity of 4 × 105 m/s enters a uniform magnetic field of 0.3 T. The velocity makes an angle of 60º with the magnetic field. Field the radius of the helical path taken by the proton beam and the pitch of the helix. The components of the proton’s velocity along and perpendicular to the magnetic field are vu = (4 × 105 m/s) cos 60º = 2 × 105 m/s. manishkumarphysics.in
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Chapter # 34
Magnetic Field
B
v1 = (4 × 105 m/s) sin 60º = 2/3 × 105 m/s. As the force qv B is perpendicular to the magnetic field, the component vn will remain constant. In the and
plane perpendicular to the field, the proton will describe a circle whose radius is obtained from the equation a circle whose radius is obtained from the equation qv1B =
or,
mv12 r
(1.67 10 27 kg) (2 3 10 5 m / s) mv 1 r= = qB (1.6 10 19 C) (0.3 T )
= 0.012 m = 1.2 cm The time taken in one complete revolution in the plane perpendicular to B is 2 3.14 0.012 m 2r T= v = . 2 3 10 5 m / s 1
The distance moved along the field during this period, i.e., the pitch
(2 10 5 m / s) 2 3.14 0.012 m =
2 3 10 5 m / s
= 0.044 m = 4.4 cm. The qualitative nature of the path of the protons is shown in figure.
z
2A 9.
y x Sol.
10.
Sol.
The magnetic moment of the loop is niA where n is the number of turns, i is the current and A is the area-vector. The direction of A is determined by the sense of the current and in this case it lies in the fourth quadrant making an angle = 37º with the negative Y-axis. Torque B niA B Thus, = 100 × (2A) × (12 × 10–4 m2) × 1 T × sin 37º = 0.14 N-m. The torque is along the positive Z-axis. An electron moves with a constant speed v along a circle of radius r. (a) Find the equivalent current through a point on its path. (b) Find the magnetic moment of the circulating electron. (c) Find the ratio of the magnetic moment of the angular momentum of the electron. (a) Consider a point P on the path of the electron. In one revolution of the electron, a charge e cross the point P. As the frequency of revolution is v/(2r), the charge crossing P in unit time, i.e., the electric current is i=
ev 2r
manishkumarphysics.in
Page # 9
Chapter # 34 Magnetic Field (b) The area A enclosed by this circular current is r2 so that the magnetic moment of the current is ev 2 evr ( r ) = iA = 2 2r in a direction perpendicular to the loop. (c) The angular momentum of the electron is = mvr. Its direction is opposite to that of the magnetic moment. Thus.
evr e 2 mvr 2 m 11.
An electron is released from the origin at a place where a uniform electric field E and a uniform magnetic field B exist along the negative Y-axis and the negative Z-axis respectively. Find the displacement of the electron along the Y-axis when its velocity becomes perpendicular to the electric field for the first time.
Y u E
F B x
Sol.
Let us take axes as shown in figure. According to the right-handed system, the Z-axis is upward in the figure and hence the magnetic field is shown downwards. At any time, the velocity of the electron may be written as u ux i uy j The electric and magnetic fields may be written as E E j and B Bk respectively. The force on the electron is = eF j eB(u y i u x j ) F e(E u B) Thus, Fx = euy B and Fy = e(E – uxB). The components of the acceleration are ax
and
ay
du x eB uy dt m
du y
dt
....(i)
e (E u xB) . m
....(ii)
We have,
d2u y dt
=–
2
eB du x m dt
=–
eB eB . u m m y
= – 2 uy
eB ....(iii) m This equation is similar to that for a simple harmonic motion. Thus, uy = A sin (t + ) .....(iv) and hence, =
where
du y dt At t = 0, uy = 0 and
du y dt
=
Fy dt
= A cos (t + )
=
....(v)
eE . m
Putting in (iv) and (v), = 0 and A =
eE E . m B
manishkumarphysics.in
Page # 10
Chapter # 34
Magnetic Field
E sin t. B The path of the electron will be perpendicular to the Y-axis when uy = 0. This will be the case for the first time at t where sin t = 0 or, wt = Thus,
uy =
or,
t=
Also,
uy =
m = eB dy E = sin t dt B
y
E
dy B sin t dt
or,
0
or,
y=
E (1 – cos t). B
At
t=
,
E 2E (1 cos ) B B Thus, the displacement along the Y-axis is
y=
2E 2Em 2Em . B BeB eB 2
EXERCISE 1.
An alpha particle is projected vertically upward with a speed of 3.0 × 10 4 k/m in a region where a magnetic field of magnitude 1.0 Tn exists in the direction south to north. Fine the magnetic force thaty acts on the particle fdlh LFkku ij pqEcdh; {ks=k 1.0 VsLyk nf{k.k ls mÙkj dh vksj gS] ;gk¡ ij ,d -d.k m/okZ/kj Åij dh vksj 3.0×104 fdeh@ls pky ls iz{ksfir fd;k x;k gSA d.k ij yxus okyk pqEcdh; cy Kkr dhft;sA Ans.
2.
An electron is projected horizontally with a kinetic energy of 10 keV. A magnetic field of strength 1.0 × 10–7 T exists in the vertically upward direction. (a) Will the electron deflect towards left of its motion? (b) Calculate the sideways deflection of the electron in travelling through 1 m. Make appropriate approximations. ,d bysDVªkWu {kSfrt fn'kk esa 10 keV ÅtkZ ls iz{ksfir fd;k x;k gSA bl LFkku ij m/okZ/kj Åij dh vksj 1.0×10–7 VsLyk rhozrk dk pqEcdh; {ks=k gS (a) D;k bysDVªkWu viuh xfr dh fn'kk ds nk;ha vFkok cka;h vksj fo{ksfir gksxk\ (b) 1 eh- nwjh r; djus esa bysDVªkWu dk fo{ksi.k Kkr dhft;sA mfpr lhek rd x.kuk dhft;sA Ans. (a) left (b) 1.5 cm A magnetic field of (4.0 × 10–3 k )T exerts a force of (4.0 i + 3.0 j ) × 10–10 N on a particle having a charge of 1.0 × 10–9 C and going in the x - y plane. Find the velocity of the particle. x-y ry esa xfr'khy] 1.0 × 10–9 dwyke vkos'k okys d.k ij (4.0 × 10–3 k ) VsLyk pqEcdh; {ks=k ds dkj.k (4.0 i + 3.0 j )
3.
× 10–10 U;wVu Ans. 4.
cy yxrk gSA d.k dk osx Kkr dhft;sA
(–75 i + 100 j ) m/s
An experimenter’s diary reads as follows; “a charged particle is projected in a magnetic field of (7.0 i – 3.0 j ) × 10–3 T. The acceleration of the particle is found to be ( i + 7.0 j ) × 10–6 m/s2n. The number to the left of
manishkumarphysics.in
Page # 11
Chapter # 34 Magnetic Field i in the last expression was not readable. What can this number be? ,d iz;ksxdrkZ dh Mk;jh esa fy[kk gqvk gS] “(7.0 i – 3.0 j ) × 10–3 VsLyk rhozrk okys pqEcdh; {ks=k esa ,d vkosf'kr d.k iz{ksfir fd;k x;k gSA d.k dk izfs {kr Roj.k ( i + 7.0 j ) × 10–6 eh/ls2 " i ds lkFk NwVh gqbZ la[;k i<+us esa ugha vk jgh gSA
;g la[;k D;k gks ldrh gS\ Ans. 6.
3.0
When a proton is released from rest in a room, it starts with an initial acceleration a0 towards west. When it is projected towards north with a speed v0, it moves with an initial acceleration 3a0 towards west. Find the electric field and the maximum possible magnetic field in the room. tc fdlh dejs esa fojkekoLFkk ls ,d bysDVªkuW eqDr fd;k tkrk gS] ;g izkjfEHkd Roj.k a0 ls if'pe dh vksj xfr'khy gksrk gSA tc bldks v0 pky ls mÙkj dh vksj iz{ksfir fd;k tkrk gSA ;g if'pe dh vksj izkjfEHkd Roj.k 3a0 ls xfr'khy gksrk gSA
dejs esa fo|qr {ks=k rFkk vf/kdre laHko pqEcdh; {ks=k Kkr dhft;sA Ans. 7.
2ma 0 ma 0 towards west, ev downward e 0
Consider a 10 cm long portion of a straight wire carrying a current of 10 A placed in a magnetic field of 0.1 T making an angle of 53º with the wire. What magnetic force does the wire experience? ,d rkj ls 10 A /kkjk izokfgr gks jgh gSA bldk 10 lseh yEck ,dne lh/kk Hkkx 0.1 VsLyk rhozrk ds pqEcdh; {ks=k ls 53º
dks.k cukrs gq, {ks=k esa j[kk gqvk gSA rkj ij pqEcdh; {ks=k ds dkj.k yxus okys cy dk ifjek.k fdruk gS\ Ans. 8.
0.08 N perpendicular to both the wire and the field
A current of 2 A enters at the corner d of a square frame abcd of side 20 cm and leaves at the opposite corner b. A magnetic field B = 0.1 T exists in the space in direction perpendicular to the plane of the frame as shown in figure. Find the magnitude and direction of the magnetic forces on the four sides of the frame. 20 lseh Hkqtk okys oxkZdkj Ýse abcd ds d dksus ls 2 A /kkjk izfo"V gksrh gS rFkk nwljs dksus b ls ckgj fudyrh gSA fp=kkuqlkj bl {ks=k esa Ýse ds ry ds yEcor~ ,d pqEcdh; {ks=k B = 0.1 VsLyk mifLFkr gSA Ýse dh pkjksa Hkqtkvksa ij pqEcdh; cy dk
ifjek.k ,oa fn'kk Kkr dhft;sA
Ans. 9.
d
c
a
b
0.02 N on each wire, on da and cb towards left and on dc and ab downward.
A magnetic field of strength 1.0 T is produced by a strong electromagnet in a cylindrical region of radius 4.0 cm as shown in figure. A wire, carrying a current of 2.0 A, is placed perpendicular to and intersecting the axis of the cylindrical region. Find the magnitude of the force acting on the wire. ,d izcy fo|qr&pqEcd dh lgk;rk ls fp=k esa n'kkZ;s vuqlkj 4.0 lseh f=kT;k okys csyukdkj {ks=k esa 1.0 VsLyk rhozrk dk
pqEcdh; {ks=k mRiUu fd;k tkrk gSA csyukdkj {ks=k dh v{k dks dkVrs gq, rFkk blds yEcor~ ,d rkj j[kk gqvk gS] ftlesa 2.0 A /kkjk izokfgr gks jgh gSA rkj ij yxus okys cy dk ifjek.k Kkr dhft;sA S S
Ans. 10.
i
0.16 N
A wire of length l carries a current i along the x-axis. Amagnetic field exists which is given as B B0 ( i j k )T . Find the magnetic of the magnetic force acting on the wire. x-v{k ds vuqfn'k j[ks gq, yEckbZ ds rkj ls i /kkjk izokfgr gks jgh gSA ogk¡ ij ,d pqEcdh; B B0 ( i j k )T }kjk O;Dr fd;k tkrk gSA rkj ij yxus okys pqEcdh; cy dk ifjek.k Kkr Ans.
11.
{ks=k mifLFkr gS] tks dhft;sA
2B 0N
A current of 5.0 A axis in the circuit shown in figure (34 – E3). The wire PQ has a lengths of 50 cm and the manishkumarphysics.in
Page # 12
Chapter # 34 Magnetic Field magnetic field in which it is immersed has a magnitude of 0.20 T. Find the magnetic force acting on the wire PQ. fp=k esa iznf'kZr ifjiFk esa 5.0 A /kkjk izokfgr gks jgh gSA rkj PQ dh yEckbZ 50 lseh gS rFkk ftl {ks=k esa ;g fLFkr gSA mldk ifjek.k 0.20 VsLyk gSA rkj PQ dh yEckbZ 50 lseh gS rFkk ftl {ks=k esa ;g fLFkr gSA mldk ifjek.k 0.20 VsLyk gSA rkj PQ ij yxus okk pqEcdh; cy Kkr dhft;sA
Ans. 12.
13.
×
×
×
×
×
× P ×
×
×
×
×
×
×
× Q ×
0.50 N towards the inside of the circuit
A circular loop of radius a, carrying a current i, is placed in a two-dimensional magnetic field. The centre of the loop cooincides with the centre of the field (figure). The strength of the magnetic field at the periphery of the loop is B. Find the magnetic force on the wire. a f=kT;k dk ,d o`Ùkkdkj ywi] ftlls i /kkjk izokfgr gks jgh gS] ,d f}&foeh; pqEcdh; {ks=k esa j[kk gqvk gSA ywi dk dsUnz pqEcdh; {ks=k ds dsUnz ls lEikfrr gS ¼fp=k½ ywi dh ifjf/k ij pqEcdh; {ks=k B gSA rkj ij pqEcdh; cy Kkr dhft;sA
Ans. 2aiB, perpendicular to the plane of the figure going into it. A hypothetical magnetic field existing in a region is given by B B 0 er , where e r denotes the unit vector along the radial direction. A circular loop of radius a, carrying a current i, is placed with its plane parallel to the X-Y plane and the centre at (0, 0, d). Find the magnitude of the magnetic force acting on the loop. fdlh {ks=k esa ,d dkYifud pqEcdh; {ks=k B B0 er mifLFkr gS] tgk¡ e r f=kT;h; fn'kk esa ,dkad lfn'k dks O;Dr djrk
gSA a f=kT;k dk ,d o`Ùkkdkj ywi] ftlls i /kkjk izokfgr gks jgh gS] x-y ry ds lekukarj bl izdkj j[kk gqvk gS fd bldk dsUnz (0, 0, d) ij gSA ywi ij yxus okys pqEcdh; cy dk ifjek.k Kkr dhft;sA Ans.
18.
2a 2iB 0 a 2 d2
2 x . A magnetic field B A wire, carrying a current, i, is kept in the X – Y plane along the curve y = A sin exists in the z-direction. Find the magnitude of the magnetic force on the portion of the wire between x = 0 and x = . 2
,d rkj ftlls i /kkjk izokfgr gS] x – y ry esa oØ y = A sin x ds vuqfn'k j[kk gqvk gSA B ifjek.k dk ,d pqEcdh; {ks=k z-fn'kk esa mifLFkr gSA x = 0 rFkk x = ds e/; fLFkr rkj ds Hkkx ij pqEcdh; cy dk ifjek.k Kkr dhft;sA
Ans. 19.
iB
A rigid wire consists of a semicircular portion of radius R and two straight sections (figure). The wire is partially immersed in a perpendicular magnetic field B as shown in the figure. Find the magnetic force on the wire if it carries a current i. ,d n`<+ rkj esa ,d R f=kT;k dk v)ZoÙ` kkdkj Hkkx rFk nks lh/ks Hkkx gS ¼fp=k½A tSlk fd fp=k esa n'kkZ;k x;k gS fd rkj ,d yEcor~ pqEcdh; {ks=k esa vkaf'kd :i ls vUnj j[kk gqvk gSA ;fn rkj ls i /kkjk izokfgr gks jgh gS rks bl ij pqEcdh; cy
Kkr dhft;sA
×
×
×
×
×
×
×
×
×
×
× × i
R
×
×
×
×
×
×
×
×
×
×
×
×
manishkumarphysics.in
Page # 13
Chapter # 34 Ans. 2iRB, upward in the figure 21.
Magnetic Field
Figure shows a rod PQ of length 20.0 cm and mass 200 g suspended through a fixed point O by two threads of lengths 20.0 cm each. A magnetic field of strength 0.500 T exists in the vicinity of the wire PQ as shown in the figure. The wires connecting PQ will the battery are loose and exert no force on PQ. (a) Find the tension in the threads when the switch S is open. (b) A current of 2.0 A is established when the switch S is closed. Find the tension in the threads now. fp=k esa n'kkZ;k x;k gS fd 20.0 lseh yEckbZ rFkk 200 xzke nzO;eku dh NM+ nks /kkxksa dh lgk;rk ls ,d fLFkj fcUnq O ls yVdk;h x;h gS] izR;sd /kkxs dh yEckbZ 20.0 lseh gSA rkj ds lehi okys {ks=k esa 0.500 VsLyk rhozrk dk ,d pqEcdh; {ks=k fp=kkuqlkj mifLFkr gSA PQ dks cSVjh ls tksM+us okys la;kstu rkj
Q
S
Ans. 22.
(a) 1.13 N (b) 1.25 N
Two metal strips, each of length l, are clamped parallel to each other on a horizontal floor with a separation b between them. A wire of mass m lies on them perependicularly as shown in figure. A vertically upward the wire and the floor is . A current i is established when the switch S is closed at the instant t = 0. Discuss the motion of the wire after the switch is closed. How far away from the strips will the wire reach? nks /kkrq dh ifV~V;k¡] izR;sd dh yEckbZ gS] {kSfrt Q'kZ ij b nwjh ij ijLij lekukarj DysEi dh x;h gSA bu ij fp=kkuqlkj m nzO;eku dk ,d rkj yEcor~ j[kk gqvk gSA vkdk'k esa m/okZ/kj Åij dh vksj B rhozrk dk pqEcdh; {ks=k mifLFkr gSA /kkrq dh ifV~V;k¡ fpduh gS] fdUrq rkj ,oa Q'kZ ds chp ?k"kZ.k&xq.kkad dk eku gSA t = 0 ij dqath yxkdj i /kkjk LFkkfir
dh tkrh gSA dqath yxkus ds i'pkr~ rkj dh xfr dh foospuk dhft;sA rkj] ifV~V;ksa ls fdruh nwj rd pyk tk;sxk\
Ans. 23.
A metal wire PQ of mass 10 g lies at rest on two horizontal metal rails separated by 4.90 cm (figure). A vertically downward magnetic field of magnitude 0.800 T exists in the space. The resistance of the circuit is slowly decreased and it is found that when the resistance goes below 20.0 , the wire PQ starts sliding on the rails. Find the coefficient of friction. 10 xzke nzO;eku okyk /kkrq dk ,d rkj 4.90 lseh nwj fLFkr /kkrq dh nks {kSfrt iVfj;ksa ij j[kk gqvk gS ¼fp=k½A vkdk'k esa m/okZ/kj uhps dh vksj 0.800 VsLyk rhozrk dk pqEcdh; {ks=k mifLFkr gSA ifjiFk dk izfrjks/k /khjs&/khjs de fd;k tkrk gS rFkk tc izfrjks/k 20.0 ls de gksrk gS] rkj PQ iVfj;ksa ij fQlyuk izkjEHk dj nsrk gSA ?k"kZ.k xq.kkad Kkr dhft;sA
Ans. 24.
×
P×
×
×
×
×
×
×
×
×
×
×
×
Q×
×
×
6V
0.12
A straight wire of length can slide on tow parallel plastic rails kept in a horizontal plane with a separation d. The coefficient of friction between the wire and the rails is . If the wire carries a current i, what minimum magnetic field should exist in the space in order to slide the wire on the rails.
manishkumarphysics.in
Page # 14
Chapter # 34
Magnetic Field
{kSfrt ry esa d nwjh ij fLFkr IykfLVd dh nks iVfj;ksa ij yEckbZ dk ,d {kSfrt rkj fQly ldrk gSA rkj rFkk iVfj;ksa ds e/; ?k"kZ.k xq.kkad gSA ;fn rkj esa izokfgr /kkjk i gS] vkdk'k esa U;wure fdruk pqEcdh; {ks=k mifLFkr gksuk pkfg;s] rkfd iVfj;ksa ij rkj fQly ldsA mg
Ans. 25.
Figure shows a circular wire-loop of radius a, carrying a current i, placed in a perependicular magnetic field B. (a) Consider a small part dl of the wire. Find the force on this part of wire exerted by the magnetic field. (b) Find the force of compression in the wire. a f=kT;k dk rkj dk ,d o`Ùkkdkj ywi] ftlls i /kkjk izokfgr gks jgh gS] fp=kkuqlkj ,d yEcor~ pqEcdh; {ks=k B esa j[kk gqvk gSA (a) rkj ds ,d vYika'k d ij fopkj dhft;sA bl Hkkx ij pqEcdh; {ks=k ds dkj.k yxus okyk cy Kkr dhft;sA (b) rkj esa lEihMu cy Kkr dhft;sA
Ans. 26.
i 1 2
(a) idlB towards the centre (b) iaB
Suppose that the radius of cross-section of the wire used in the previous problem is r. Find the increase in the radius of the loop if the magnetic field is switched off. The Young’s modulus of the material of the wire is Y. ekuk fd fiNys iz'u esa iz;D q r rkj ds vuqiLz Fk dkV dh f=kT;k r gSA ;fn pqEcdh; {ks=k fLop vkWQ dj fn;k tk;s rks ywi dh f=kT;k esa o`f) Kkr dhft;sA rkj ds inkFkZ dk ;ax dk izR;kLFkrk xq.kkad Y gSA
ia 2B Ans. 27.
r 2Y
.
The magnetic field existing in a region is given by
fdlh {ks=k esa mifLFkr pqEcdh; {ks=k fuEukuqlkj O;Dr fd;k tkrk gS % x B = B0 1 kˆ
A square loop of edge and carrying a current i, is placed with its edges parallel to the x-y axes. Find the magnitude of the net magnetic force experienced by the loop. Hkqtk dk ,d oxkZdkj ywi ftlesa i /kkjk izokfgr gS] bldh Hkqtk,¡ x-y v{kksa ds lekukarj j[krs gq, j[kk x;k gSA ywi }kjk
vuqHkfor dqy pqEcdh; cy dk ifjek.k Kkr dhft;sA Sol.
The magnitude of magnetic field varies along x-axis. It can be observed that force on wire PQ and RS have same magnitude but opposite directions. or FPQ + FRS = 0
Net magnitude of force on square loop is = FSP FQR a a = – i B0 1 + i B0 1
28.
= i B0
A conducting wire of length l, lying normal to a magnetic field B, moves with a velocity v as shown in figure. (a) Find the average magnetic force on a free electron of the wire. (b) Due to this magnetic force, electrons manishkumarphysics.in
Page # 15
Chapter # 34 Magnetic Field concentrate at one end resulting in an electric field inside the wire. The redistribution stops when the electric force on the free electrons balances the magnetic force. Find the electric field developed inside the wire when the redistribution stops. (c) What potential difference is developed between the ends of the wire? pqEcdh; {ks=k B ds vfHkyEcor~ j[kk gqvk yEckbZ dk pkyd rkj fp=kkuqlkj v osx ls xfr dj jgk gSA (a) rkj ds eqDr bysDVªkuW ij vkSlr pqEcdh; cy Kkr dhft;sA (b) bl cy ds dkj.k bysDVªkWu ,d fljs ij ,df=kr gks tkrs gSa rFkk rkj ds vUnj
,d fo|qr {ks=k mRiUu gks tkrk gS] ;g iqufoZrj.k rc :drk gSA tc eqDr bysDVªkWu ij oS|rq cy] pqEcdh; cy ls larfq yr gks tkrk gSA tc iqufoZrj.k :d gkrk gSA rkj esa fo|qr {ks=k Kkr dhft;sA (c) rkj ds fljksa ds e/; fdruk foHkokarj mRiUu gksrk gSA × ×
× × v
× × × × Ans. 29.
(a) evB (b) vB
×
× ×
(c) iBv
A current i is passed through a silver strip of width d and area of cross-section A. The number of free electrons per unit volume is n. (a) Find the drift velocity v of the electrons. (b) If a magnetic field B exists in the region as shown in figure, what is the average magnetic force on the free electrons? (c) Due to the magnetic force, the free electrons get accumulated on one side of the conductor along its length. This produces a transverse electric field in the conductor which opposes the magnetic force on the electrons. Find the magnitude of the electric field which will stop further accumulation of electrons. (d) What will be the potential difference developed across the width of the conductor due to the electron accumulation? The appearance of a transverse emf, when a current-carrying wire is placed in a magnetic field, is called Hall effect. d pkSM+kbZ o A vuqiLz Fk&dkV {ks=kQy okyh pkanh dh iV~Vh ls i /kkjk izokfgr gSA izfr ,dkad vk;ru esa eqDr bysDVªkWuksa dh la[;k n gSA (a) bysDVªkWuksa viokg osx v Kkr dhft;sA (b) ;fn pqEcdh; {ks=k fp=k esa iznf'kZr {ks=k esa gh mifLFkr gS] eqDr bysDVªkWu ij vkSlr pqEcdh; cy fdruk gS\ (c) pqEcdh; cy ds dkj.k] eqDr bysDVªkWu rkj dh yEckbZ ds vuqfn'k blds
,d fljs ij ,df=kr gks tkrs gSAa blds dkj.k pkyd esa vuqiLz Fk fo|qr {ks=k mRiUu gks tkrk gS] tks bysDVªkWuksa dk ,df=kdj.k jksd nsAa (d) bysDVªkWuksa ds ,df=kdj.k ds dkj.k pkyd dh pkSM+kbZ ds e/; fdruk foHkokUrj mRiUu gksxk\ pqEcdh; {ks=k esa j[ks gq, /kkjkokgh rkj esa bl izdkj mRiUu vuqizLFk fo|qr okgd cy] gkWy&izHkko dgykrk gSA ×
×
× ×
× i
× Ans. 30.
(a)
i Ane
(b)
× ×
v
×
×
×
×
×
× d
×
×
×
×
×
×
×
×
×
×
iB upwards in the figure An
(c)
iB Ane
(d)
iBd Ane
A particle having a charge of 2.0 × 10–8 C and a mass of 2.0 × 10–10 g is projected with a speed of 2.0 × 103 m/s in a region having a uniform magnetic field of 0.10 T. The velocity is pependicular to the field. Find the radius of the circle formed by the particle and also the time period. 0.10 VsLyk rhozrk okys pqEcdh; {ks=k esa 2.0 × 10–8 dwykWe rFkk 2.0 × 10–10 xzke nzO;eku okyk ,d d.k 2.0 × 103 eh/ls
pky ls iz{ksfir fd;k x;k gSA osx dh fn'kk {ks=k ds yEcor~ gSA d.k ds }kjk r; fd;s x;s o`Ùkkdkj iFk dh f=kT;k rFkk vkorZdky Kkr dhft;sA Ans.
20 cm, 6.3 × 10–4 s
31.
A proton describes a circle of radius 1 cm in a magnetic field of strength 0.10 T. What would be the radius of the circle described by an -particle moving with the same speed in the same magnetic field? 0.10 VsLyk rhozrk okys pqEcdh; {ks=k esa ,d izkVs hu 1 lseh f=kT;k dk o`Ùkkdkj iFk r; djrk gSA blh {ks=k esa leku pky ls xfr'khy -d.k }kjk r; fd;s x;s o`Ùkkdkj iFk dh f=kT;k fdruh gksxh\ Ans. 2 cm
32.
An electron having a kinetic energy of 100 eV circulates in a path of radius 10 cm in a magnetic field. Find the magnetic field and the number of revolutions per second made by the electron. fdlh pqEcdh; {ks=k esa 100 eV xfrt&ÅtkZ okyk bysDVªkuW 10 lseh f=kT;k dk o`Ùkkdkj iFk r; djrk gSA pqEcdh; {ks=k rFkk
bysDVªkWu }kjk izfr lsd.M yxk;s x;s pDdjksa dh la[;k Kkr dhft;sA Ans.
3.4 × 10–4 T, 9.4 × 10–6 manishkumarphysics.in
Page # 16
Chapter # 34 Magnetic Field 33. Protons having kinetic energy K emerge from an acceleration as a narrow beam. The beam is bent by a perependicular magnetic field so that it just misses a plane target kept at a distance l in front of the accelerator. Find the magnetic field. ,d Rofj=k ls K xfrt ÅtkZ ds izkVs kWu ladjs iqt a ds :i esa mRlftZr gksrs gSAa ,d yEcor~ pqEcdh; {ks=k }kjk iat q dks bruk eksM+k tkrk gS fd ;g Rofj=k ls nwjh ij j[ks y{; ls Vdjkus ls cl pwd tkrk gSA pqEcdh; {ks=k Kkr dhft;sA Ans. 34.
2mpK el
A charged particle is accelerated through a potential difference of 12 kV and acquires a speed of 1.0 × 10–6 m/s. It is then injected perpendicularly into a magnetic field of strength 0.2 T. Find the radius of the circle described by it. 12 kV foHkokarj ls Rofjr fd;k x;k ,d vkosf'kr d.k 1.0 × 10–6 eh/ls pky izkIr dj ysrk gSA blds i'pkr~ bls 0.2 VsLyk
rhozrk okys pqEcdh; {ks=k esa izfo"V djk;k tkrk gSA blds }kjk cuk;s x;s o`Ùk dh f=kT;k Kkr dhft;sA Ans. 36.
12 cm
A proton is projected with a velocity of 3 × 106 m/s perpendicular to a uniform magnetic field of 0.6T. Find the acceleration of the proton. 0.6 VsLyk rhozrk ds pqEcdh; {ks=k ds yEcor~ ,d izkVs kWu 3 × 106 eh/ls pky ls iz{ksfir fd;k tkrk gSA izkVs kWu dk Roj.k Kkr
dhft;sA Ans. 37.
(a) 3.2 × 10–15 N (b) 2.1 × 10–4 m (c) 1.31 × 10–7 s
(a) An electron moves along a circle of radius 1 m in a perpendicular magnetic field of strength 0.50 T. What would be its speed? Is it reasonable? (b) If a proton moves along a circle of the same radius in the same magnetic field, what would be its speed? (a) 0.50 T rhozrk okys yEcor~ pqEcdh; {ks=k esa ,d bysDVªkuW 1 eh- f=kT;k ds o`Ùkkdkj iFk esa Hkze.k djrk gSA bldh pky fdruh gksxh\ D;k ;g laHko gSA (b) ;fn leku pqEcdh; {ks=k esa leku f=kT;k okys o`Ùkkdkj iFk ij izkVs kWu izfr'khy gks rks
bldh pky fdruh gksxh\ Ans. 38.
(a) 8.8 × 10 10 m/s (b) 4.8 × 10 7 s
A particle of mass m and positive charge q, moving with a uniform velocity v, enters a magnetic field B as shown in figure. (a) Find the radius of the circular are it describes in the magnetic field. (b) Find the angle subtended by the are at the centre. (c) How long does the particle stay inside the magnetic field? (d) Solve the three parts of the above problem if the charge q on the particle is negative. ,d leku osx v ls xfr'khy m nzO;eku rFkk q /kukRed vkos'k okyk ,d d.k] fp=kkuqlkj pqEcdh; {ks=k B esa izfo"V gksrk gSA (a) pqEcdh; {ks=k esa blds }kjk cuk;s x;s o`Ùkkdj pki dh f=kT;k Kkr dhft;sA (b) pki }kjk dsUnz ij varfjr dks.k Kkr dhft;sA (c) d.k pqEcdh; {ks=k esa fdrus le; rd jgsxkA (d) ;fn d.k ij _.kkos'k gS] rks mDr rhuksa Hkkx gy dhft;sA v
× × × Ans. 40.
(a)
× × ×
mv (b) – 2 qB
× × ×
× × ×B × × × (c)
m ( – 2) qB
(d)
mv m , + 2, ( – 2) qB qB
A narrow beam of singly-charged carbon ions, moving at a constant velocity of 6.0 × 104 m/s, is sent perpendicularly in a rectangular region having uniform magnetic field B = 0.5 T (figure). It is found that two beams emerge from the field in the backward direction, the separations from the incident beam being 3.0 cm and 3.5 cm. Identify the isotopes present in the ion beam. Take the mass of an ion = A(1.6 × 10–27) kg, where A is the mass number. ,d vk;rkdkj izn's k esa fLFkr B = 0.5 T pqEcdh; {ks=k es]a fu;r osx 6.0 × 104 eh/ls, xfr'khy ,d/kk vk;fur dkcZu vk;uksa dk iat q ] yEcor~ izfo"V djk;k tkrk gS ¼fp=k½A ;g ik;k x;k fd ihNs dh fn'kk esa nks iat q ls nwfj;k¡ 3.0 lseh rFkk 3.5 lseh. gSA vk;u iat q ls mifLFkr vk;uksa dh igpku dhft;sA ,d vk;u dk nzO;eku = A(1.6 × 10–27) fdxzk-] tgk¡ A nzO;eku la[;k
gSA
manishkumarphysics.in
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Chapter # 34
Ans.
Magnetic Field
12
× × ×
× × ×
× × ×
×
×
×
14
C and C
41.
Fe+ ions are accelerated through a potential difference of 500 V and are injected normally into a homogeneous magnetic field B of strength 20.0 mT. Find the radius of the circular paths followed by the isotopes with mass numbers 57 and 58. Take the mass of an ion = A(1.6 × 10–27) kg, where A is the mass number. 500 oksYV foHkokarj ls Rofjr Fe+ vk;u 20.0mT rhozrk okys le:i pqEcdh; {ks=k esa vfHkyEcor~ izfo"V djk;s tkrs gSA nzO;eku la[;k 57 rFkk 58 okys leLFkkfudksa }kjk r; fd;s x;s o`Ùkkdkj iFkksa dh f=kT;k Kkr dhft;sA ,d vk;u dk nzO;eku = A(1.6 × 10–27) fdxzk eku yhft;s] tgk¡ A nzO;eku la[;k gSA Ans. 119 cm and 120 cm
43.
Figure shows a convex lens of focal length 12 cm lying in a uniform magnetic field B of magnitude 1.2 T parallel to its principal axis. A particle having a charge 2.0 × 10–3 C and mass 2.0 × 10–6 kg is projected perpendicular to the plane of the diagram with a speed of 4.8 m/s. The particle moves along a circle with its centre on the principal axis at a distance of 18 cm from the lens. Show that the image of the particle goes along a circle and find the radius of that circle. fp=k esa n'kkZ;k x;k gS fd 12 lseh f=kT;k okyk ,d mÙky ySla ] ,d le:i pqEcdh; {ks=k B ftldk ifjek.k 1.2 T rFkk fn'kk bldh eq[; v{k ds vuqfn'k gS] j[kk gqvk gSA 2.0 × 10–3 dwykWe vkos'k rFkk 2.0 × 10–6 fdxzk nzO;eku okyk ,d d.k fp=k ds yEcor~ 4.8 eh/ls pky ls iz{ksfir fd;k x;k gSA d.k ,d o`Ùkkdkj iFk esa xfr djrk gS ftldk dsUnz ySl a ls 18 lseh nwj
gSA O;Dr dhft;s fd d.k dk izfrfcEc o`Ùkkdkj iFk ij xfr djsxk rFkk bl o`Ùk dh f=kT;k Kkr dhft;sA B P
Ans. 44.
8 cm
Electrons emitted with negligible speed from an electron gun are acceleration through a potential difference V along the x-axis. These electrons emerges from a narrow hole into a uniform magnetic field B directed along this axis. However, some of the electrons emerging from the hole make slightly divergent angles as shown in figure. Show that these paraxial electrons are refocused on the X-axis at a distance ,d bysDVªkWu&xu ls bysDVªkWu ux.; pky ls mRlftZr gksrs gSa rFkk x-v{k ds vuqfn'k V foHkokarj ls Rofjr fd;s tkrs gSAa ;g bysDVªkWu ,d fNnz ls le:i pqEcdh; {ks=k B esa izo's k djrs gS]a ftldh fn'kk bl v{k ds vuqfn'k gSA ;|fir mRlftZr gksus okys dqy bysDVªkWu fp=kkuqlkj dqN vilkjh dks.k cukrs gSAa O;Dr dhft;s fd v{k ds lehi okys ;g bysDVªkuW x-v{k
ij fuEu nwjh ij iqu% Qksdflr gks tk;sxsaA 8 2mV eB 2
Ans. 45.
Two particles, each having a mass m are placed at a separation d in a uniform magnetic field B as shown in figure. They have opposite charges of equal magnitude q. At time t = 0, the particles are projected towards each other, each with a speed v. Suppose the Coulomb force between the charges is switched off. (a) Find the maximum value vm of the projection speed so that the two particles do not collide. (b) What would be the minimum and maximum separation between the particles if v = vm/2? (c) At what instant will a collision occur between the particle if v = 2vm? (d) Suppose v = 2vm and the collision between the particles is completely inelastic. Describe the motion after the collision. fp=kkuqlkj ,d le:i pqEcdh; {ks=k B es]a nks d.k izR;sd dk nzO;eku m, d nwjh ij j[ks gq, gSaA bu ij leku ifjek.k q ds foijhr vkos'k gSA le; t = 0 ij] d.k ,d nwljs dh vksj izR;sd v pky ls iz{ksfir fd;s tkrs gSAa ekukfd d.kksa ds chp manishkumarphysics.in
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Chapter # 34
Magnetic Field
dqykWeh; cy lekIr gks x;k gSA (a) iz{ksi.k pky dk og vf/kdre eku vm Kkr dhft;s fd d.k vkil esa u Vdjk;sA (b) ;fn v = vm/2 gS rks d.kksa ds chp U;wure rFkk vf/kdre nwfj;k¡ fdruh gS? (c) ;fn v = 2vm gks rks d.kksa ds chp VDdj fdl {k.k ij gksxh\ (d) ekuk yhft;s fd v = 2vm gS rFkk d.kksa ds chp VDdj iw.kZr;k izR;kLFk gSA VDdj ds i'pkr~ xfr dh foospuk dhft;sA
m qBd q 3q (b) , (c) 6 qB 2m 2 2 (d) the particles stick together and the combined mass moves with constant speed vm along the straight line drawn upward in the plane of figure through the point of collision. Ans.
46.
(a)
A uniform magnetic field of magnitude 0.20 T exists in space from east to west. With what speed should a particle of mass 0.010 g and having a charge 1.0 × 10–5 C be projected from south to north so that it moves with a uniform velocity? vkdk'k esa iwoZ ls if'pe dh vksj bafxr 0.20 T dk le:i pqEcdh; {ks=k mifLFkr gSA 0.010 xzke nzO;eku rFkk 1.0×10–5 dwykWe vkos'k okyk ,d d.k nf{k.k ls mÙkj dh vksj fdl pky ls iz{ksfir fd;k tk;s fd ;g ,d leku osx ls
xfr djs\ Ans.
47.
49 m/s
A particle moves in a circle of diameter 1.0 cm under the action of a magnetic field of 0.40 T. An electric field of 200 V/m makes the path straight. Find the charge/mass ratio of the particle. 0.40 T rhozrk okys pqEcdh; {ks=k ds izHkko esa ,d d.k 1.0 lseh O;kl okys o`Ùkkdkj iFk ij xfr djrk gSA 200 oksYV@eh
dk fo|qr {ks=k blds iFk dks lh/kk dj nsrk gSA d.k ds vkos'k@nzO;eku dk vuqikr Kkr dhft;sA Ans.
48.
2.5 × 10 5 C/kg
A proton goes underflected in a crossed electric and magnetic field (the field are perependicular to each other) at a speed of 2.0 × 106 m/s. The velocity is perependicular to both the fields. When the electric field is switched off, the proton moves along a circle of radius 4.0 cm. Find the magnitude of the electric and the magnetic fields. Take the mass of the proton = 1.6 × 10–27 kg. ,d Økflr fo|qr ,oa pqEcdh; {ks=k esa ¼{ks=k ijLij yEcor~ gS½] 2.0 × 106 eh/ls pky ls xfr'khy izkVs kWu fcuk fo{ksfir gq, xfr djrk gSA osx dh fn'kk nksuksa {ks=kksa ds yEcor~ gSA tc fo|qr {ks=k fLop vkWQ dj fn;k tkrk gS] izkVs kWu 4.0 lseh f=kT;k
okys o`Ùkkdkj iFk ij xfr djrk gSA fo|qr {ks=k rFkk pqEcdh; {ks=k ds ifjek.k Kkr dhft;sAs izksVkWu dk nzO;eku = 1.6 × 10–27 fdxzk eku yhft;sA Ans.
49.
1.0 × 104 m/s, 1.0 × 105
A particle having a charge of 5.0 C and a mass of 5.0 × 10–12 kg is projected with a speed of 1.0km/s 1.0 km/ s in a magnetic field of magnitude 5.0 mT. The angle between the magnetic field and the velocity is sin–1 (0.90). Show that the path of the particle will be a helix. Find the diameter of the helix and its pitch. ,d d.k ftl ij vkos'k 5.0 C rFkk nzO;eku 5.0 × 10–12 fdxzk gS] 1.0 fdeh@ls pkys ls 5.0 mT ifjek.k ds pqEcdh; {ks=k esa iz{ksfir fd;k tkrk gSA pqEcdh; {ks=k rFkk osx ds chp dks.k sin–1 (0.90) gSA O;Dr dhft;s fd d.k dk iFk gsfyDl gksxkA
gsfyDl dk O;kl rFkk bldh fip Kkr dhft;sA Ans.
36 cm, 56 cm
50.
A proton projected in a magnetic field of 0.020 T travels along a helical path of radius 5.0 cm and pitch 20 cm. Find the components of the velocity of the proton along and perpendicular to the magnetic field. Take the mass of the proton = 1.6 × 10–27 kg. 0.020 T rhozrk okys pqEcdh; {ks=k esa iz{s ksfir fd;k x;k ,d izkVs kWu gsfydy iFk ds vuqfn'k xfr djrk gS] ftldh f=kT;k 5.0 lseh rFkk fip 20 lseh gSA izkVs kWu ds osx ds pqEcdh; {ks=k ds vuqfn'k ,oa yEcor~ ?kVd Kkr dhft;sA izksVkWu dk nzO;eku = 1.6 × 10–27 fdxzk eku yhft;sA Ans. 6.4 × 10 4 m/s, 1.0 × 10 5
51.
A particle having mass m and charge q is released from the origin in a region in which electric field and magnetic field are given by B B0 j and E E0k . Find the speed of the particle as a function of its z-coordinate. manishkumarphysics.in
Page # 19
Chapter # 34 m nzO;eku
gS %
Magnetic Field
rFkk q vkos'k okyk ,d d.k ,d ls {ks=k es]a ewy fcUnq ls NksMk+ tkrk gS] tgk¡ pqEcdh; {ks=k rFkk fo|qr {ks=k fuEukuqlkj B B0 j rFkk E E0k .
d.k dh pky] blds z-funs'Z kkad ds Qyu ds :i esa Kkr dhft;sA Ans. 53.
2qE 0 z m
A rectangular coil of 100 turns has length 5 cm and width 4 cm. It is placed with its plane parallel to a uniform magnetic field and a current of 2A is sent through the coil. Find the magnitude of the magnetic field B, if the torque acting on the coil is 0.2 N-m. ,d vk;rkdkj dq.Myh dh yEckbZ 5 lseh rFkk pkSM+kbZ 4 lseh gSA bldks ,d le:i pqEcdh; {ks=k esa bl izdkj j[kk x;k gS fd bldk ry pqEcdh; {ks=k ds lekukarj gS rFkk dq.Myh esa 2A /kkjk izokfgr dh tk jgh gSA ;fn dq.Myh ij 0.2 U;wVu&eh-
cy dk ifjek.k Kkr dhft;sA Ans.
54.
0.5 T
A 50-turn circular coil of radius 2.0 cm carrying a current of 5.0 A is rotated in a magnetic field of strength 0.20 T. (a) What is the maximum torque that acts on the coil? (b) IN a particular position of the coil, the torque acting on it is half of this maximum. What is the angle between the magnetic field and the plane of the coil? ,d 50 Qsjs okyh rFkk 2.0 lseh f=kT;k okyh o`Ùkkdkj dq.Myh esa 5.0 A /kkjk izokfgr gks jgh gS] bldks 0.20 T rhozrk ds pqEcdh; {ks=k esa ?kqek;k tkrk gSA (a) dq.Myh ij yxus okyk vf/kdre cy vk?kw.kZ fdruk gS\ (b) dq.Myh dh ,d fof'k"V fLFkfr es]a
bl ij yx jgk cy vk?kw.kZ] vf/kdre dk vk/kk gksrk gS] dq.Myh ds ry rFkk pqEcdh; {ks=k ds chp dks.k fdruk gksxk\ Ans.
55.
(a) 6.3 × 10–2 N-m
(b) 90º
A rectangular loop of sides 20 cm and 10 cm carries a current of 5.0 A. A uniform magnetic field of magnitude 0.20 T exists parallel to the longer side of the loop. (a) What is the force acting on the loop? (b) What is the torque acting on the loop? ,d vk;rkdkj ywi dh Hkqtk,¡ 20 lseh ,oa 10 lseh gS] blls 5.0 A /kkjk izokfgr gks jgh gSA 0.20 T rhozrk dk ,d le:i pqEcdh; {ks=k mifLFkr gSA ftldh fn'kk bldh yEch Hkqtk ds lekukarj gSA (a) ywi ij yx jgk cy fdruk gS\ (b) ywi ij
yx jgk cy vk?kw.kZ fdruk gS\ Ans.
(a) zero (b) 0.02 N-m parallel to the shorter side.
56.
A circular coil of radius 2.0 cm has 500 turns in it and carries a current of 1.0 A. Its axis makes an angle of 30º with the uniform magnetic field to magnitude 0.40 T that exists in the space. Find the torque acting on the coil. ,d o`Ùkkdkj ywi dh f=kT;k 2.0 lseh rFkk 500 Qsjs gS] blesa 1.0 A /kkjk izokfgr gks jgh gSA bldh v{k 0.40 T rhozrk ds le:i pqEcdh; {ks=k ls 30º dks.k cukrh gSA dq.Myh ij yxus okyk cy vk?kw.kZ Kkr dhft;sA Ans. 0.13 N-m
57.
A circular loop carrying a current i has wire of total length L. A uniform magnetic field B exists parallel to the plane of the loop. (a) Find the torque on the loop. (b) If the same length of the wire is used to form a square loop, what would be the torque? Which is larger? ,d o`Ùkkdkj ywi ls i /kkjk izokfgr gks jgh gS rFkk blds rkj dh dqy yEckbZ L gSA ywi ds ry ds lekukarj ,d le:i pqEcdh; {ks=k B mifLFkr gSA (a) ywi ij cy vk?kw.kZ Kkr dhft;sA (b) ;fn bruh gh yEckbZ ds rkj ls ,d oxkZdkj ywi cuk;k tk;s]
cy&vk?kw.kZ fdruk gksxk\ dkSulk vf/kd gS\ Ans.
58.
(a)
iL2B 4
(b)
iL2B 16
A square coil of edge having n turns carries a current i. It is kept on a smooth horizontal plate. A uniform magnetic field B exists in a direction parallel to an edge. The total mass of the coil is M. What should be the minimum value of B for which the coil will starts tipping over? ,d oxkZdkj ywi dh Hkqtk rFkk Qsjksa dh la[;k n gS] blesa i /kkjk izokfgr gks jgh gSA ;g ,d fpduh {kSfrt ifV~Vdk ij
j[kk gqvk gSA ,d le:i pqEcdh; {ks=k ftldh fn'kk bldh ,d Hkqtk ds vuqfn'k gS] mifLFkr gSA dq.Myh dk dqy nzO;eku M gSA B dk U;wure eku fdruk gks fd dq.Myh ,d fdukjs ls frjNh gksdj Åij mBus yxs\ Ans.
Mg 2 nil
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Chapter # 34 Magnetic Field 59. Consider a nonconducting ring of radius r and mass m which has a total charge q distributed uniformly on it. The ring is rotated about its axis with an angular speed . (a) Find the equivalent electric current
q in the ring. (b) Find the magnetic of the ring. (c) Show that = 2 m where is the angular momentum of the ring about its axis of rotation. ,d vpkyd o`Ùkkdkj oy; dh f=kT;k r rFkk nzO;eku m gS] bl viuh v{k ds ifjr% dks.kh; pky ls ?kqek;k tkrk gSA (a) dq.Myh Kkr
q
dhft;sA (c) O;Dr dhft;s fd = 2 m , tgk¡ dq.Myh dk bldh v{k ds ifjr% dks.kh; laosx gSA
Ans. 60.
ij dqy q vkos'k ,d leku :i ls forfjr gSA cy dks esa rqY; /kkjk Kkr dhft;sA (b) dq.Myh dk pqEcdh; vk?kw.kZ
(a)
q 2
(b)
qr 2 2
Consider a nonconducting plate of radius r and mass m which has a charge q distributed uniformly over it. The plate is rotated about its axis with an angular speed . Show that the magnetic moment m and
q the angular momentum of the plate are related as = 2 m .
,d dqpkyd ifV~Vdk dh f=kT;k r rFkk nzO;eku m gS] bl ij q vkos'k ,d leku :i ls forfjr gSA ifV~Vdk dks bldh v{k ds ifjr% dks.kh; pky ls ?kqek;k tkrk gSA O;Dr dhft;s fd ifV~vdk dks dks.kh; laoxs rFkk pqEcdh; vk?kw.kZ esa q
ijLij lEcU/k = 2 m gSA 61.
Consider a solid sphere of radius r and mass m which has a charge q distributed uniformly over its volume. speed . Show that the magnetic moment and the angular momentum of the sphere are
q related as 2 m . r f=kT;k
ds ,d Bksl xksys dk nzO;eku m gS rFkk bl ij q vkos'k blds vk;ru esa ,d leku :i ls forfjr gSA xksys dks blds O;kl ds ifjr% dks.kh; pky ls ?kqek;k tkrk gSA O;Dr dhft;s fd xksys dk dks.kh; laoxs rFkk pqEcdh; vk?kw.kZ esa lac/a k
q gSA 2m
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