A uniform magnetic field is restricted within a region of radius “r”. The magnetic field changes with time at a rate dB→dt. Loop 1 of radius R > r encloses the region r and loop 2 of radius R is outside the region of magnetic field as shown in the figure. Then the e.m.f. generated is

- d B → d t π r 2 in loop 1 and zero in loop 2

A uniform magnetic field is restricted within a region of radius “r”. The magnetic field changes with time at a rate dB→dt. Loop 1 of radius R > r encloses the region r and loop 2 of radius R is outside the region of magnetic field as shown in the figure. Then the e.m.f. generated is

- d B → d t π r 2 in loop 1 and zero in loop 2

- d B → d t π r 2 in loop 1 and - d B → d t π r 2 in loop 2

- d B → d t π R 2 in loop 1 and zero in loop 2

- d B → d t π r 2 in loop 1 and zero in loop 2

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A uniform magnetic field is restricted within a region of radius “r”. The magnetic field changes with time at a rate $\frac{d \vec{B}}{d t}$ . Loop 1 of radius R > r encloses the region r and loop 2 of radius R is outside the region of magnetic field as shown in the figure. Then the e.m.f. generated is

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$- \frac{d \vec{B}}{d t} π r^{2} in loop 1 and - \frac{d \vec{B}}{d t} π r^{2} in loop 2$

$zero in loop 1 and zero in loop 2$

$- \frac{d \vec{B}}{d t} π R^{2} in loop 1 and zero in loop 2$

$- \frac{d \vec{B}}{d t} π r^{2} in loop 1 and zero in loop 2$

answer is D.

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Detailed Solution

Emf generated in loop 1,

$\begin{array}{l} ε_{1} = - \frac{d ϕ}{d t} = - \frac{d}{d t} (\vec{B} \cdot \vec{A}) = - \frac{d}{d t} (B A) = - A \times \frac{d B}{d t} \\ ε_{1} = - (π r^{2} \frac{d B}{d t}) \end{array}$

$(∵ A = π r^{2} because \frac{d B}{d t} is restricted upto radius r .)$

Emf generated in loop 2,

$ε_{2} = - \frac{d}{d t} (B A) = - \frac{d}{d t} (0 \times A) = 0$

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