A wheel with 10 metallic spokes each 0.5m long is rotated with a speed of 120 rev/min in a plane normal to the earth's magnetic field at the place. If the magnitude of the field is 0.4 G, the induced e.m.f. between the axle and the rim of the wheel equal to

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1.256 × 10^–3 V

6.28 × 10^–4 V

1.256 × 10^–4 V

6.28 × 10^–5 V

answer is D.

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Detailed Solution

Given data:

Metallic spoke of length r=0.5m

Frequency $f = 120 \frac{r e v}{\min} = 2 \frac{r e v}{\sec}$

Magnetic field $B = 0.4 G = 0.4 \times 10^{- 4} T$

Angular velocity $ω = 2 π f = 2 π \times 2 = 4 π$

Concept used: Induced EMF

Detailed solution:

Induced EMF is given by,

$E = \frac{1}{2} {Br}^{2} ω$

$E = \frac{2 π \times 2 \times 0.4 \times 10^{- 4} \times (0 . 5^{2})}{2}$

$E = 6.28 \times 10^{- 5} V$

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