CBSE Class 12 · 2008 · All India · Electromagnetic InductionLong Answer

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Explain briefly, with the help of a labelled diagram, the basic principle of the working of an AC generator. In an AC generator, a coil of

N

turns and area

A

is rotated at

ν

revolutions per second in a uniform magnetic field

B

. Write the expression for the emf produced. A

100

-turn coil of area

0.1 m^{2}

rotates at half a revolution per second. It is placed in a magnetic field of

0.01 T

perpendicular to the axis of rotation of the coil. Calculate the maximum voltage generated in the coil.

Answer: $ε = N B A (2 π ν) sin (2 π ν t)$ ; $ε_{0} = N B A (2 π ν) \approx 0.314 V$ .

Solution

Principle: when a coil is rotated in a uniform magnetic field about an axis perpendicular to the field, the flux linked with it changes and an emf is induced (electromagnetic induction).
As the coil of

N

turns and area

A

rotates at

ν

rev/s,

ω = 2 π ν

, and the flux linkage is

N ϕ = N B A cos ω t

. The induced emf is

ε = - \frac{d ( N ϕ )}{d t} = N B A ω sin ω t = ε_{0} sin ω t

with peak value

ε_{0} = N B A ω = N B A (2 π ν)

.
Numerical:

N = 100

A = 0.1 m^{2}

B = 0.01 T

ν = 0.5 rev/s

ε_{0} = 100 \times 0.01 \times 0.1 \times 2 π \times 0.5 = 0.1 π

ε_{0} \approx 0.314 V

CBSE Class 12 Board · 2008 · All India · Subjective

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CBSE Class 12 · 2008 · All India · Electromagnetic InductionLong Answer

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