(a) Draw a labelled diagram of an a.c. generator and derive the expression for the instantaneous value of the emf induced in the coil. (b) A circular coil of cross-sectional area $200\ \text{cm}^{2}$ and 20 turns is rotated about its vertical diameter with an angular speed of $50\ \text{rad s}^{-1}$ in a uniform magnetic field of magnitude $3.0\times10^{-2}\ \text{T}$. Calculate the maximum value of the e.m.f. induced in the coil.

(a) $e=NBA\omega\sin\omega t$; (b) $e_0=0.6$ V (600 mV).

CBSE Class 12 · 2017 · Delhi · Alternating CurrentLong Answer

(a) Draw a labelled diagram of an a.c. generator and derive the expression for the instantaneous value of the emf induced in the coil.
(b) A circular coil of cross-sectional area

200 cm^{2}

and 20 turns is rotated about its vertical diameter with an angular speed of

50 rad s^{- 1}

in a uniform magnetic field of magnitude

3.0 \times 1 0^{- 2} T

. Calculate the maximum value of the e.m.f. induced in the coil.

Answer: (a) $e = N B A ω sin ω t$ ;
(b) $e_{0} = 0.6$ V (600 mV).

Solution

(a) A coil of

N

turns and area

A

rotates at angular speed

ω

in a field

B

. The flux is

ϕ = N B A cos ω t

, so

e = - \frac{d ϕ}{d t} = N B A ω sin ω t = e_{0} sin ω t, e_{0} = N B A ω

(b) With

N = 20

A = 200 cm^{2} = 2 \times 1 0^{- 2} m^{2}

ω = 50 rad s^{- 1}

and

B = 3.0 \times 1 0^{- 2} T

e_{0} = N B A ω = 20 \times (3.0 \times 1 0^{- 2}) \times (2 \times 1 0^{- 2}) \times 50

e_{0} = 0.6 V = 600 mV

e_{0} = 0.6 V

CBSE Class 12 Board · 2017 · Delhi · Subjective

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CBSE Class 12 · 2017 · Delhi · Alternating CurrentLong Answer

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