Suggested+Solution+FyBCh20 21NVC05+Magnetic+Induction+&+AC+Circuits

Suggested solution FyBCh20-21 NVC05 magnetic Induction & AC circuits

NV-College

Part I: Answer part I on this paper 1.

The figure below the magnetic field lines between two repelling north magnetic poles are illustrated. At which point is the magnetic field largest? [1/0] a) A C b) B c) C D d) D Why? Explain. [0/1] Answer: Alternative C. The magnetic field lines are more closed pack at point C. The strength of the magnetic field at a given point is in general illustrated by the density of the magnetic field lines at the point.

2.

+ + + + + + + + + + + + In the figure below an electron is projected into the region between two parallel charged plates which are 5.0 mm apart. The potential difference between the plates is − − − − − − − − − − − − 500. V . Data: d = 5.0 mm = 5.0 × 10 −3 m , V = 500 . V a) Draw the electric filed lines as clear as possible on the figure.

Suggested answer: The electric field is uniform in the region between the plate. Its direction is from positive to negative, as illustrated in the figure. b)

e

[1/0]

+ + + + + + + + + + + +

− − − − − − − − − − − −

In which direction will the electron be deflected? i) Toward the bottom of the paper. ii) Into the page. iii) Toward the top of the page. iv) Out of the page.

[1/0]

Answer: Alternative: iii. The electron will be deflected towards the top of the paper. This may be explain either by using the fact that the negative charges, and therefore electrons, are attracted to the r r r positive charges. We may also use F = QE = −eE to explain the effect, i.e. in the region between the plates; there is a force on the incoming electrons in the opposite direction of the electric field.

© [email protected] ☺Free to use for educational purposes.

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c)

What is the magnitude of the electric field? [1/0] i) 100 . mV / m ii) 100. kV / m iii) 2500 V / m iv) 100 . V / m Suggested solution: Answer: Alternative ii: E = 100 kV / m −3 Data: d = 5.0 mm = 5.0 × 10 m , V = 500 . V V 500 . V V = E⋅d ⇔ E = ⇔ E = = 100 × 10 +3 V / m = 100 kV / m −3 d 5.0 × 10 m

d)

The magnitude of the electric force on the electron is: [1/0] −14 i) 1.6 × 10 . N ii) 1.6 × 10 −19. N iii) 1.6 × 10 −17 . N iv) 4.0 × 10 −16. N Suggested solution: Answer: Alternative i: F = 1.6 × 10 −14 N Data: d = 5.0 mm = 5.0 × 10 −3 m , V = 500 . V , E = 100 kV / m r r r F = Q ⋅ E = −eE ⇔ F = 1.6 × 10 −19 × 100 × 10 3 = 1.6 × 10 −14 N

e)

A proton moves from the positively charged plate to the negatively charged plate. The force on the proton [1/0] i) remains the same. ii) decreases. iii) increases. iv) decreases first, reaches to an equilibrium and then rises. Suggested solution: Answer: Alternative i: F = 1.6 × 10 −14 N Data: d = 5.0 mm = 5.0 × 10 −3 m , V = 500 . V , E = 100 kV / m The force on a proton that moves from the positively charged plate to the negatively charged plate, in the region between the plate, and not so close to the edges, is constant. This is due to the fact that the electric field between two relatively close oppositely charged plate is relatively uniform and constant: r r r F = Q ⋅ E = eE ⇔ F = 1.6 × 10 −19 × 100 × 10 3 = 1.6 × 10 −14 N


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3.

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An α particle (nucleus of He atom) is shot into the region between the poles of a magnet. The α particle has a charge of + 2 ⋅ e and initially is moving at 4.0 km / s perpendicular to the direction of the magnetic field. The magnetic field is uniform and its magnitude is 0.30 T Data: Q = +2 ⋅ e , v = 4.0 km / s = 4.0 × 10 3 m / s , B = 0.30 T a) The α particle will [1/0] i) move straight and will not be deflected by the + 2e magnetic field. ii) will be deflected toward the left. iii) will be effected toward the right. iv) will be effected into the page. v) will be effected out of the page. Suggested solution: Answer: Alternative v The alpha particle will experience a magnetic force which is out of the paper and towards the readers of these words. Therefore, the alpha particle will be deflected out of the page and toward the reader. The magnitude of the force is: r r r F = Q⋅v × B ⇔ F = QVB sin 90 = 2 × 1.6 × 10 −19 × 4.0 × 10 3 × 0.30 = 3.84 × 10 −16 N b)

The magnitude of the magnetic force on the α particle is: [1/0] −16 i) 3.84 × 10 . N ii) 3.84 × 10 −19. N iii) 1.6 × 10 −19. N iv) 1.6 × 10 −16. N Suggested solution: Answer: Alternative i: F = 3.84 × 10 −16 N The alpha particle will experience a magnetic force which is out of the paper and towards the readers of these words. Therefore, the alpha particle will be deflected out of the page and toward the reader. The magnitude of the force is: r r r F = Q⋅v × B ⇔ F = QVB sin 90 = 2 × 1.6 × 10 −19 × 4.0 × 10 3 × 0.30 = 3.84 × 10 −16 N

c)

If the polarity of the magnet is reversed, the magnitude of the force [1/0] i) remains the same. ii) decreases. iii) increases. iv) decreases first, reaches to an equilibrium and then rises. Suggested solution: Answer: Alternative i: F = 3.84 × 10 −16 N If the polarity of the magnet is reversed, the magnitude of the force on the alpha particle will remain the same; only its direction will be reversed: i.e. the alpha particle will move into the page and away from the readers of these statements: The magnitude of the force is: r r r F = Q⋅v × B ⇔ F = QVB sin 90 = 2 × 1.6 × 10 −19 × 4.0 × 10 3 × 0.30 = 3.84 × 10 −16 N


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d)

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If the α particle is replaced by an electron, compare to the magnitude of the force on the α particle, the magnetic force on the electron [1/0] i) remains the same. ii) decreases. iii) increases. iv) decreases first, reaches to an equilibrium and then rises. Suggested solution: Answer: Alternative ii: F = 1.92 × 10 −16 N If the α particle is replaced by an electron, compare to the magnitude of the force on the α particle, the magnetic force on the electron will decrease. The direction of the force on the electron will opposite to that of the protons. This is due to the fact that the charge of electrons is negative. The magnitude of the force is: r r r F = Q⋅v × B ⇔ F = QVB sin 90 = 1.6 × 10 −19 × 4.0 × 10 3 × 0.30 F = 1.92 × 10 −16 N

4.

The magnetic flux through a coil of wire containing five loops change from 44 Wb to − 46 Wb in 0.18 s . What is the emf induced in the coil? [2/0] Suggested solution: Data: N = 5 , Φ B1 = 44 Wb , Φ B 2 = −46 Wb , Δt = 0.18 s , Problem: ε = ? Answer: The emf induced in the coil is ε = −3.5 kV . Φ B2 − Φ B1 dΦ B − 46 − (44 ) ⇔ ε = −N ⇔ ε = −5 × ε = −N = +2500 V = +2.5 kV 0.18 dt Δt The positive sign demonstrates the fact that the induced emf opposes the change (decrease) in the magnetic flux (Lenses law.)

5.

The two wires of a 6.5 m long appliance cord are 2.0 mm apart and carry a current of 3.0 A dc. Calculate the magnitude and direction of the force between the wires. (2/0) Suggested solution: r = 2.0 mm = 2.0 × 10 m Data: I 1 = I 2 = 3.0 A , l = 6.5 m , I = 3 .0 A −3

1

−3

r = 2.0 mm = 2.0 × 10 m, Problem: F = ? Answer: The wires repel each other with the repulsive force of F21 = 5.85 mN , F12 = 5.85 mN

B 2 = 300 μ T F12 = 5 .85 mN

B1 = 300 μ T F21 = 5 .85 mN

μ 0 I1 μ0 I ⎧ ⎧ μ 0 I1 ⎪B = ⎪ B1 = ⇔ ⎨ r 2rπ r r ⇔ ⎨ r 2π r r r ⇔ F21 = I 2 l 2π r ⎪F = I l × B ⎪F = I l × B I = 3 .0 A ⎩ 2 1 ⎩ 21 F21 F12 μ 0 I1 I 2 repulsive = = l l 2π r Due to the fact that the currents are in the opposite directions they repel each other. 2 ( 4/ π/ × 10 −7 3 .0 ) F21 = F12 = × × 6.5 F21 = F12 = 5.85 × 10 −3 N ≈ 5.9 mN −3 2/ π/ 2/ .0 × 10 Second method: The magnetic field of the first wire at the other one: 2


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μ 0 I1 4/ π/ × 10 −7 3 .0 ⇔ B1 = × = 3.0 × 10 − 4 T = 300 μT −3 2/ π/ 2/ .0 × 10 2π r B1 = 300 μT , B2 = 300 μT r r r F21 = I 2 l × B1 ⇔ F21 = 3.0 × 6.5 × 3.0 × 10 −4 = 5.85 × 10 −3 N ≈ 5.9 mN F21 = 5.85 mN Similarly B1 = 300 μT and F21 = 5.85 mN , F12 = 5.85 mN B1 =

6.

What is the direction of the induced current in the circular loop due to the current shown? Why? Explain and Draw it on the figure: [1/1] Explanation: The direction of the induced current in the ring is clock-wise. This is due to the fact that the direction of the magnetic field due to the wire inside the ring is into the paper, but because the current in the wire decreases, its magnetic field also decreases, and therefore, the magnetic flux through the ring decreases too. Therefore, there will be an induced emf in the circular loop and due to the Lens’s law the direction of the induced current must oppose the decrease in the magnetic flux.

I decreasing

I decreasing


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7.

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The figures below illustrate magnetic fields in different directions. In each figure, there is a circular loop that may be rotated about an axis AB shown also on the figure. In which situation (s) a current may be induced in the loop? Why explain. [2/2] Suggested solutions: a. Answer: The rotation axis AB is perpendicular to the magnetic field. Therefore, it is possible to produce a changing magnetic flux to induce current in the loop.

A

B

B

b. Answer: The rotation axis AB is perpendicular to the magnetic field. Therefore, it is possible to produce a changing magnetic flux to induce current in the loop.

A B

c. Answer: The rotation axis AB is perpendicular to the magnetic field. Therefore, it is possible to produce a changing magnetic flux to induce current in the loop. A

d. Answer: The rotation axis AB is parallel to the magnetic field. Therefore, the magnetic flux through the loop is always zero, and therefore, there can never be any induce current in the loop.

B

A

Suggested solutions: Answer: alternatives a, b and c. In all of the cases above, we may start with assumption that the loop has got an initial angular velocity, i.e. it is rotating, and therefore, the magnetic flux, if any, changes.

(

)

r r dΦ B d B⋅A d (BA cos θ ) dθ ε =− =− =− = BA sin θ = BA ω sin (ωt ) dt dt dt dt In the situation (d) the magnetic field is parallel to the surface of the loop and therefore magnetic flux through the loop is always zero and therefore no current may be induced in the loop. In the alternatives a, b and c, if the loop has any initial angular velocity, i.e. if the loop is rotating initially, the angle between the plane of the loop and the magnetic field changes that results in the change in the magnetic flux, and therefore, there will be a changing magnetic flux.


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8.

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An electron and a proton have the same kinetic energy upon entering a region of constant magnetic field. What is the ratio of the radii of their circular paths? [1/2] r me rP Suggested solution: Answer: e = ; ≈ 43 rb mP re Data: E KEe = E KE P , e = 1.6 × 10 −19 C , me = 9.10938188 × 10 -31 kg ≈ 9.11 × 10 -31 kg , m P = 1.67262158 × 10 -27 kg = 1.67 × 10 -27 kg r Problem: e = ? rP Answer: r r r F = Q⋅v × B F = e⋅v⋅B ⎧F = e ⋅ v ⋅ B v 2/ ⎪ 2 ⇔ e⋅v⋅B = m⋅ ⇔ / ⎨ v r ⎪ FC = ma C = m ⋅ r ⎩ v r = m⋅ e⋅B Assuming that both the electron and proton were originally moving at the same initial kinetic energy in an inertial frame, i.e.: 1 1 me ve2 = m p v 2p 2/ 2/ 2 mp mp ve v ⇔ e = = 2 v p me vp me

v e⋅B me ⋅ v e r m ⋅v m v ⇔ e = e/ ⋅ B/ = e e = e ⋅ e rb mP ⋅ v p mP ⋅ v p m P v p e/ ⋅ B/ mp re me me = ⋅ = rb mP me mP

ve

−e

B

vp

r = m⋅

Answer:

re = rb

vp

+e

me mP

r rP 1.67262158 × 10 -27 kg Answer: P ≈ 43 = 1836.15 = 42.85 ≈ 43 = -31 re 9.10938188 × 10 re kg Note that due to their opposite charges, protons and electrons path are in two opposite directions as illustrated above.


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9.

The armature of a 50 Hz ac generator rotates in a 0.60 T magnetic field. If the radius of the coil is 5.0 cm . How many loops must the coil contain if the peak output is to be 120 V ? [1/3] Suggested solution: Data: f = 50 Hz , B = 0.60 T , r = 5.0 cm = 0.050 m , ε 0 = 120 V Problem: N = ? Answer: The generator must have N = 81 loops in order to produce ε 0 = 120 V . ε ε0 ε = NBA ω sin (ωt ) ⇔ ε 0 = NBA ω ⇔ N = 0 ⇔ N = ⇔ 2 BAω π ⋅ r B ⋅ 2π ⋅ f ε0 120 ⇔ N= N = 81 = 81.06 ≈ 81 N= 2 2 2 2π × 0.05 2 × 0.60 × 50 2π ⋅ r B ⋅ f

10.

An RC circuit has a resistor of resistance 15.0 kΩ and a capacitor of capacitance 3.00 μF . The capacitor is at voltage V0 initially. How long does it take the capacitor to discharge to 1.0% of its initial voltage? [1/3] Suggested Solutions: Data R = 15 .0 kΩ = 15 .0 × 10 3 Ω , C = 3.00 μF = 3.00 × 10 −6 F , Problem: t = ? if V (t ) = 0.01V0

V = V0 e

−

t RC

15 . 0 k Ω

V0

3 . 00 μ F

; τ = RC ⇔

τ = (15.0 × 10 3 Ω )(3.00 × 10 −6 F ) τ = 45.0 × 10 −3 s

1,2

45.0×10− 3 s

0.01V/ 0 = V/ 0 e Taking the natural logarithm of both sides, and using ln e = 1 leads to: t ln(0.01) = − 45.0 ×10 −3 s

t = −45.0 ×10 −3 ⋅ ln(0.01) = 0.207 s

V = V0e

−

t RC

R = 15 .0 kΩ

t

1

C = 3 .00 μF

0,8 V (volt)

−

Discarging a capacitor

τ = RC = 45.0 ×10 −3 s

0,6 0,4 0,2 0 0

0,05

0,1

0,15

0,2

0,25

t (s)

Answer: t = 207 ms


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0,3


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11. If 75 MW of power at 55 kV arrives at a town from a generator via 5.0 Ω transmission lines, calculate a. the emf at the generator end of the lines, the fraction of the power generated that is lost in the lines. b. Suggested solution: Answer: ε ≈ 62 kV ; ΔPdissipated % ≈ 11 %

(2/1) (2/1)

Data: P = 75 MW , V = 55 kV , R = 5.0 Ω Problem: ε = ? , ΔPlost % = ? Answer: The emf of the generator is ε ≈ 62 kV . ΔPdissipated % ≈ 11 % of the power is dissipated in the wires during the transmission of the power from the generator to the town. P P 75 × 10 6 P = IV ⇔ I = ⇔ I = = ≈ 1364 A ⇔ V V 55 × 10 3 The emf of the generator may be calculated as: ε = V + RI ⇔ ε = V + RI = 55 × 10 3 + 5 × 1364 = 6.18 × 10 4 V

I ≈ 1364 A

Answer: ε ≈ 62 kV

The power dissipated in the wire may be calculated as: 2 Pdissipated = RI 2 ⇔ Pdissipated = 5.0 × (1364 ) = 9297521 ≈ 9.3 × 10 6 W = 9.3 MW Pdissipated ≈ 9.3 × 10 6 W Total power at the power plant: Ptot = Ptown + Pdissipated ≈ 75 × 10 6 + 9.3 × 10 6 W ≈ 84.3 × 10 6 W 9297521 × 100 ≈ 0.11 = 11% 84.3 × 10 6 Ptot 12% of the power is dissipated in the wires during the transmission of the power from the generator to the town: ΔPdissipated % ≈ 11 % ΔPdissipated % =

Pdissipated

× 100 ⇔ ΔPdissipated % =


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12. Two stiff parallel wires a distance l apart in a horizontal plane act as rails to support a light metal rod of mass m that is parallel to the surface of the ground and perpendicular to the parallel rails as illustrated NORTH below. A uniform magnetic field B directed vertically downward (into the page in the figure) acts WEST EAST throughout. The system is connected to a electric power source that provides the system with a constant current I . SOUTH a)

In which direction does the rod move? Why? Explain! i) ii) iii) iv) v) vi)

East. West North. South. Up. Down.

[1/1]

NORTH

WEST

r r r F = Il × B

B

EAST

Suggested solution: Answer: I Il B SOUTH Alternative: ii: West; v = ⋅t ; m IlB − μ k mg v = at = ⋅t m r r r The magnetic force on a current I in a magnetic field B is F = I l × B . Using the right hand rule we may recognize that there is a force to the left, i.e. to the west on the mobile rod, and therefore, the rod moves to the left. It accelerates to the left at the rate of Il B . a= m Calculate the speed of the rod as a function of time

[2/4/¤]

if the rails are considered frictionless. if the coefficient of the friction between the rod and the trails is μ k . IlB − μ k mg Il B Suggested solution: Answer: v = ⋅ t ; v = at = ⋅t m m Il B on a frictionless surface, As discussed above, the rod accelerates at the rate of a = m and if the initial velocity of the rod is assumed zero, its velocity as a function of time may be expressed as: Il B v = v0 + at ⇒ v = ⋅t m If the coefficient of the kinetic friction between the rod and the rail is μ k , the friction force may be expressed as F f = μ k FN = μ k mg and it is in the opposite direction of the

b) c)

motion and the resultant force on the rod may be expressed as: © [email protected] ☺Free to use for educational purposes.

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13.

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⎧ F = FB − F f = IlB − μ k mg IlB − μ k mg IlB − μ k mg ⇔a= ⇔ v = at = ⋅t ⎨ m m ⎩ F = ma A mass spectrometer is being used to monitor air pollutants. It is difficult, however, to separate molecules with nearly equal mass such as CO ( 28.0106 u ) and N 2 ( 28.0134 u ). How large a radius of curvature must a spectrometer have if these two molecules are to be separated on film by 1.00 mm . (2/4/¤) Suggested solution: Data: mCO = 28 .0106 u , m N 2 = 28.0134 u Δd = 2 rCO − 2 rN 2 = 1.00 mm = 1.00 × 10 −3 m ,

Problem: r = ? Answer: The radius of curvature of the mass spectrometer must be r ≈ 5.00 m In the mass spectroscope ions are produced by heating, or by an electric current. These ions are directed through a relatively small hole to a region known as “velocity selector” r r where there are both electric field E (pointing up) and magnetic field B1 (pointing out of the paper toward the reader.) The magnitudes of the fields are adjusted such that those E will follow a straight path passing ions whose velocity fits, Q/ E = Q/ vB1 , i.e. v = B1 r through a second hole and entering a new region where magnetic field B2 (pointing out of the paper toward the reader) exists. The ions will then follow a circular path of radius v2 r satisfying QvB 2 = m . The radius r may be measured using a photographic plate r where ions strike and darken the film: E E ⎧ v= ⎧QE = QvB1 ⎪ B v E E B1 ⎪ ⎪ ⇔ QB 2 = m = m 1 = m ⇔ r=m ⎨ v2 ⇔ ⎨ r r rB1 QB1 B2 ⎪QvB2 = m ⎪QB = m v r ⎩ 2 r ⎩⎪ Due to the fact that ions are assumed having the same chare (for example ionized once) r r r sharing identical electric and magnetic fields E , B1 and B2 we may express the radius E ⎧ r m = N N 2 2 ⎪ QB1 B2 ⎪ of curvature of each beam as ⎨ E ⎪r = m CO CO ⎪⎩ QB1 B2 rN mN2 Dividing the equations to each other may result in: 2 = ⇔ rCO mCO mN2 ⎛ mN ⎞ Δd rN 2 = rCO ⇔ Δd = 2(rN 2 − rCO ) = 2⎜⎜ 2 − 1⎟⎟ rCO ⇔ rCO = mCO ⎛ mN ⎞ ⎝ mCO ⎠ 2⎜⎜ 2 − 1⎟⎟ ⎝ mCO ⎠ −3 1.00 × 10 m r ≈ rCO = = 5.00189 m ⇔ r ≈ rCO = 5.00189 m ≈ 5.00 m ⎛ 28.0134 u ⎞ − 1⎟⎟ 2⎜⎜ ⎝ 28.0106 u ⎠


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Suggested+Solution+FyBCh20 21NVC05+Magnetic+Induction+&+AC+Circuits

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