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 below. Then the e.m.f. generated is

Emf⁡(e)=−dϕdt;ϕ=BA=πr2B;dϕdt=−πr2dBdte1=−dϕ1dt=−πr2dBdt;e2=−dϕ2dt=0∵ϕ2=0

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 below. Then the e.m.f. generated is

Emf⁡(e)=−dϕdt;ϕ=BA=πr2B;dϕdt=−πr2dBdte1=−dϕ1dt=−πr2dBdt;e2=−dϕ2dt=0∵ϕ2=0

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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 $\frac{\vec{dB}}{dt}$ . 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 below. Then the e.m.f. generated is

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Zero in loop 1 and zero in loop 2

$- \frac{dB}{dt} {πr}^{2}$ in loop 1 and $- \frac{d \vec{B}}{dt} {πr}^{2}$ in loop 2

$- \frac{d \vec{B}}{dt} {πR}^{2}$ in loop 1 and zero in loop 2

$- \frac{dB}{dt} {πr}^{2}$ in loop 1 and zero in loop 2

answer is D.

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Detailed Solution

$\begin{array}{l} Emf (e) = \frac{- dϕ}{dt}; ϕ = BA = {πr}^{2} B; \frac{dϕ}{dt} = - {πr}^{2} \frac{dB}{dt} \\ e_{1} = \frac{- {dϕ}_{1}}{dt} = - {πr}^{2} \frac{dB}{dt}; e_{2} = \frac{- {dϕ}_{2}}{dt} = 0 [∵ ϕ_{2} = 0] \end{array}$

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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 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 below. Then the e.m.f. generated is

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