(C) The magnetic flux $\phi$ through the loop is given by $\phi = B \cdot A = 20 \times 10^{-2} \times 2 \sin(50 \pi t) = 0.4 \sin(50 \pi t) \,Wb$. Th

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(C) The magnetic flux $\phi$ through the loop is given by $\phi = B \cdot A = 20 \times 10^{-2} \times 2 \sin(50 \pi t) = 0.4 \sin(50 \pi t) \,Wb$. The induced current is $I = \frac{e}{R} = -\frac{1}{R} \frac{d\phi}{dt}$. The charge flowing through the loop is $\Delta q = \int_{0}^{t} I \,dt = -\frac{1}{R} \int_{\phi_1}^{\phi_2} d\phi = -\frac{\phi_2 - \phi_1}{R}$. At $t = 0$, $\phi_1 = 0.4 \sin(0) = 0 \,Wb$. At $t = 20 \,ms = 0.02 \,s$, $\phi_2 = 0.4 \sin(50 \pi \times 0.02) = 0.4 \sin(\pi) = 0 \,Wb$. Thus, the net charge $\Delta q = -\frac{0 - 0}{20} = 0 \,C$.

Class 12 Physics Electromagnetic Induction Mix Examples-Electromagnetic Induction

Medium

$A$ conducting circular loop of resistance $20 \Omega$ and cross-sectional area $20 \times 10^{-2} \,m^{2}$ is placed perpendicular to a spatially uniform magnetic field $B$, which varies with time $t$ as $B = 2 \sin(50 \pi t) \,T$. Find the net charge flowing through the loop in $20 \,ms$ starting from $t = 0$. (in $\,C$)

WBJEE 2020 All WBJEE

A
$0.5$
B
$0.2$
C
$0$
D
$0.14$

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The dimension of the ratio of magnetic flux and the resistance is equal to that of:

Shown in the figure is a circular loop of radius $r$ and resistance $R$. $A$ variable magnetic field of induction $B = B_0 e^{-t}$ is established inside the coil. If the key $K$ is closed,the electrical power developed right after closing the switch is equal to

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