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A thin metallic ring of radius 5 cm is made of a wire of resistance $2 \frac{Ω}{m}$ . The ring is placed in a uniform magnetic field of induction 1 Tesla with its plane perpendicular to the field. Now the ring starts rotating with angular velocity $ω = 10 r a d / \sec$ . Then rms value of current induced in the ring is

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$\frac{1}{4 \sqrt{2}} A$

$\frac{1}{8 \sqrt{2}} A$

$\frac{1}{2 \sqrt{2}} A$

$4 \sqrt{2} A$

answer is B.

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Detailed Solution

At any time t, $ϕ = B A \cos θ = π r^{2} B \cos ω t$

$∴ ε = - \frac{d ϕ}{d t} = π r^{2} B ω \sin ω t$

$∴ i = \frac{ε}{R} = \frac{π r^{2} B ω \sin ω t}{2 π r . λ} = \frac{r B ω}{2 λ} \sin ω t$

$∴ I_{m} = \frac{r B ω}{2 λ} a n d I_{r m s} = \frac{I_{m}}{\sqrt{2}} = \frac{0.05 \times 1 \times 10}{2 \times 2 \times \sqrt{2}} A = \frac{1}{8 \sqrt{2}} A$

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