Solutions for Exercises of Electromagnetic Induction - Grad Plus

# Solutions for Exercises of Electromagnetic Induction

1) In a Faraday disc dynamo, a metal disc of radius R rotates with an angular velocity perpendicular to the plane of the disc and passing through its centre. The disc is placed in a magnetic field B acting perpendicular to the plane of the disc. Determine the induced emf between the rim and the axis of the disc.

Consider a metal disc of radius R rotating with ω and placed in normal magnetic field B as shown in figure
Consider infinitesimal width of the disc of length equal to radius R. This can be considered as a rod of length L = R rotating in normal magnetic field B with angular speed ω.
We know that emf induced between the ends of the rod of length L normal magnetic field B with angular speed ω is given by,

$latex i.e.\;e=\frac12BL^2\omega=\frac12BR^2\omega$

2) A horizontal wire 20 m long extending from east to west is falling with a velocity of 10 m/s normal to the Earth’s magnetic field of 0.5 x10-4 T. What is the value of induced emf in the wire?

L = 20m, v = 10m/s, B = 0.5 × 10-4T, e = ?

i.e. e = BLv

∴ e = 0.5 × 10-4 × 20 × 10

= 10 × 10-3V

3) A metal disc is made to spin at 20 revolutions per second about an axis passing through its centre and normal to its plane. The disc has a radius of 30 cm and spins in a uniform magnetic field of 0.20 T, which is parallel to the axis of rotation. Calculate (a) The area swept out per second by the radius of the disc, (b) The flux cut per second by a radius of the disc, (c) The induced emf in the disc.

f = 20Hz, r = 30cm = 30 × 10-2m, B = 0.20 T,

A = ? swept by radius in one second

ϕ = ? cut by radius in one second

e = ?

The area swept out by the radius of the disc is,

$latex A\;=\;\frac12r^2\theta$

$latex i.e.\;A\;=\;\frac12r^2\left(\omega t\right)$

$latex i.e.\;A\;=\;\frac12r^2\left(2\pi ft\right)$

$latex =\;r^2\pi ft$

$latex \therefore\;A\;=\;{(30\;\times\;10^{-2})}^2\pi\;\times\;20\;\times\;$

$latex \therefore\;A\;=\;5.65m^2$

The flux cut by a radius of the disc is given as,

ϕ = AB

Hence flux cut in one second is,

∴ ϕ = 5.65 × 0.20 = 1.13Wb

emf induced in the disc is given as,

$latex e=\frac12Br^2\omega$

$latex =\frac12\;\times\;0.2\;\times\;\left(30\times10^{-2}\right)^2\;\times\;2\pi\;\times\;20\;$

∴ e = 1.13V

4) A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 10 A in 0.2 s, what is the change of flux linkage with the other coil?

M = 1.5, dI = (10 − 0) = 10A, dt = 0.2s,

$latex \frac{d\phi}{dt}=e_{sec}=?\;(other\;coil)$

We know that,

$latex e_{sec}=M\frac{dI_p}{dt}$

$latex \therefore\;e_{sec}\;=\;1.5\;\times\;\frac{10}{0.2}$

$latex \therefore\;e_{sec}\;=\;75V$

Hence change of flux per second with other coil is 75Wb.

5) A long solenoid consisting of 1.5 x103 turns/m has an area of cross-section of 25 cm2. A coil C, consisting of 150 turns (Nc) is wound tightly around the centre of the solenoid. Calculate for a current of 3.0 A in the solenoid (a) the magnetic flux density at the centre of the solenoid, (b) the flux linkage in the coil C, (c) the average emf induced in coil C if the current in the solenoid is reversed in direction in a time of 0.5 s.

n = 1.5 × 103, A = 25cm2 = 25 × 10-4m2,

NC = 150, I = 3 A, B (inside solenoid) =?, ϕC =?,

eC =? for current in solenoid reversed in 0.5s

a) Magnetic flux density, B, at the centre is given as,

B = μ0nI

= 4π × 10-7 × 1.5 × 103 × 3

∴ B = 5.654× 10-3T

b) Flux linked with coil C is given as,

ϕC = NCAB

= 150 × 25 × 10-4 × 5.654× 10-3

∴ ϕC = 2.12 × 10-3Wb

c) When current through coil 1 is reversed, change in magnetic field is given as,

dB = (5.654 × 10-3) − (−5.654× 10-3)

As current is reversed direction of B is reversed.

∴ dB = 11.3 × 10-3T

$latex \therefore\;e_C\;=\frac{d\phi_C}{dt}=\frac{d\left(N_CAB\right)}{dt}$

$latex \therefore\;e_C\;=N_CA\frac{dB}{dt}$

$latex \therefore\;e_C\;=\;150\;\times\;25\;\times\;10^{-4}\;\times\;\frac{11.3\times10^{-3}}{0.5}$

$latex \therefore\;e_C\;=\;8.48\;\times\;10^{-3}V\;$

6) A search coil having 2000 turns with area 1.5 cm2 is placed in a magnetic field of 0.60T. The coil is moved rapidly out of the field in a time of 0.2 second. Calculate the induced emf across the search coil.

N = 2000, A = 1.5cm2= 1.5 × 10-4m2, B = 0.60T,

t = 0.2 s, e = ?

As coil is moved completely out of the field in t = 0.2 s,

dA = (1.5 × 10-4 − 0)

= 1.5 × 10-4m2,

also dt = 0.2s

$latex \therefore\;e\;=\;\frac{d\phi}{dt}$

$latex \therefore\;e\;=\;\frac{d\left(NAB\right)}{dt}=\;NB\frac{dA}{dt}$

$latex \therefore\;e\;=\;2000\;\times\;0.60\;\times\frac{1.5\times10^{-4}}{0.2}$

$latex \therefore\;e\;=\;0.9\;V$

7) An aircraft of wing span of 50 m flies horizontally in earth’s magnetic field of 6×10-5T at a speed of 400 m/s. Calculate the emf generated between the tips of the wings of the aircraft.

L = 50m, B = 6 × 10-5T, v = 400m/s, e = ?

e = BLv

∴ e = 6 × 10 × 50 × 400

∴ e = 1.2V

8) A stiff semi-circular wire of radius R is rotated in a uniform magnetic field B about an axis passing through its ends. If the frequency of rotation of wire be f, calculate the amplitude of alternating emf induced in the wire

Flux passing through the coil is given as,

ϕ = ABcosθ.

Here θ is the angle between area vector and field.

$latex \therefore\;\phi\;=\;AB\cos\theta\;=\frac{\pi R^2}2\times\;B\;\times\;\cos\left(2\pi ft\right)$

Hence emf induced in the coil is given by,

$latex \therefore e\;=\;-\frac{d\phi}{dt}=\frac d{dt}\left(\frac{\pi R^2}2\times B\times\cos\left(2\pi ft\right)\right)$

$latex \therefore e\;=\;\frac{\pi R^2}2\times B\times\sin\left(2\pi ft\right)\times\left(2\pi f\right)$

$latex \therefore\;e\;=\;\pi^2BR^2f\sin(2\pi ft)\;=\;e_0\sin(2\pi ft)\;$

Hence peak value is (π2BR2f).

9) Calculate the value of induced emf between the ends of an axle of a railway carriage 1.75 m long travelling on level ground with a uniform velocity of 50 km per hour. The vertical component of Earth’s magnetic field (Bv) is given to be 5 x 10-5T.

L = 1.75m, B = 5 × 10-5T,

$latex v\;=\;50\frac{km}{hr}=50\times\frac5{18}$

$latex i.e.\;v\;=\;13.89\;\frac ms,\;e=?$

we know that,

e = BvLv

∴ e = 5 × 10-5 × 1.75 × 13.89

∴ e = 1.22 × 10-3V

10) The value of mutual inductance of two coils is 10 mH. If the current in one of the coil changes from 5A to lA in 0.2 s, calculate the value of emf induced in the other coil. Also calculate the value of induced charge passing through the coil if its resistance is 5 ohm.

M = 10mH = 10-2H, dI = (5 − 1) = 4A

dt = 0.2s, esec =?, induced =? if R = 5 Ω.

We know that,$latex e_{sec}\;=\;M\frac{dI_p}{dt}$

$latex \therefore\;e_{sec}\;=\;10^{-2}\;\times\;\frac4{0.2}\;=\;0.2V$

$latex e\;=\;IR\;\;\;\;\therefore\;I\;=\;\frac eR$

$latex \therefore\;I\;=\;\frac{0.2}5=0.04\;A$

$latex But,\;I\;=\;\frac qt\;\;\;\;\;\therefore\;q\;=\;It$

$latex \;q=\;0.04\;\times\;0.2\;$

$latex \;q\;=\;8\;\times\;10^{-3}C$

11) An emf of 96.0 mV is induced in the windings of a coil when the current in a nearby coil is increasing at the rate of 1.20 A/s. What is the mutual inducta (M) of the two coils?

es = 96mV = 96 × 10-3V,

$latex \frac{dI_p}{dt}=1.20\frac As,\;M=?$

$latex e_s\;=\;M\frac{dI_p}{dt}$

$latex \therefore\;M\;=\;\frac{e_s}{\left(dI_p/dt\right)}$

$latex \therefore\;M\;=\;\frac{96\;\times\;10^{-3}}{1.20}$

$latex \therefore\;M\;=\;80\;\times\;10^{-3}H$

12) A long solenoid of length l, cross-sectional area A and having N1 turns (primary coil), has a small coil of N2 turns (secondary coil) wound about its centre. Determine the Mutual inductance (M) of the two coils.

Magnetic field due to first solenoid is,

B1 = μ0N1I1

Hence flux linked with coil-2 due to this field,

ϕ2 = N2AB1

Here A1 = A1 = A, as second coil is wound on first one. Area of cross-section will be same.

∴ ϕ2 = N20N1I1 = μN1N2AI1

$latex \therefore\;\frac{\phi_2}{I_1}=\mu_0N_1N_2A$

13) The primary and secondary coil of a transformer each have an inductance of 200 x 10-6H. The mutual inductance (M) between the windings is 4 × 10-6 H.
What percentage of the flux from one coil reaches the other?

LP = Ls = 200 × 10-6H, M = 4 × 10-6H, K=?

We have,

$latex M\;=\;k\sqrt{L_PL_s}$

$latex \therefore\;k\;=\;\frac M{\sqrt{L_pL_s}}$

$latex \therefore\;k\;=\;\frac{4\times10^{-6}}{\sqrt{\left(200\times10^{-6}\right)^2}}$

$latex \therefore\;k\;=\;\frac{4\times10^{-6}}{\left(200\times10^{-6}\right)}$

$latex \therefore\;k\;=\;0.02$

Hence flux coupled with other coil is 0.02 or 2%.

14) A toroidal ring, made from a bar of length (I) 1 m and diameter (d) 1 cm, is bent into a circle. It is wound tightly with 100 turns per cm. If the permeability of bar is equal to that of free space (μ0), Calculate the magnetic field inside the bar (B) when the current (i) circulating through the turns is 100 A. Also determine the self-inductance (L) of the coil.

???? = 1m, d = 1cm = 1 × 10-2m,

n = 100per cm = 100 × 100=104 per m, I = 100A,

μ = μ0, B = ? and L = ?

Magnetic field inside the toroid is given as,

B = μ0nI

= 4π × 107 × 104 × 100 = 1.256T

Flux linked with the coil is given as,

$latex \therefore\;\phi\;=\;NAB$

$latex \therefore\;\phi\;=\;n\;\times\;l\;\times\;\pi\;\times\;\left(\frac d2\right)^2\;\times\;B$

$latex \therefore\;\phi\;=\;10^4\times\;1\;\times\;3.14\;\times\;\left(\frac{10^{-2}}2\right)^2\times1.256$

$latex \therefore\;\phi\;=\;1Wb$

$latex \therefore\frac{\;\phi}1\;=\;\frac1{100}\;=\;10^{-2}$

But ratio of flux linked with coil due to its current is called as self inductance.

∴ L = 10-2H = 10mH

15) A uniform magnetic field B(t), pointing upward fills a circular region of radius, s in horizontal plane. If B is changing with time, find the induced electric field.

According to laws of electromagnetic induction, for

$latex e\;=\;-\frac{d\phi}{dt}$

$latex \therefore\;e\;=\;-\;\frac{d(AB)\;}{dt}$

$latex \therefore\;e\;=\;-\;A\frac{dB}{dt}$

$latex \therefore\;e\;=\;-\;\pi s^2\frac{dB}{dt}$

Hence induced electric field is proportional to $latex -\pi s^2\frac{dB}{dt}$

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