A square loop of side 10 cm and resistance 0 | vedclass

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(C) The area vector $\vec{A}$ of the loop is perpendicular to the east-west plane,which is in the north-south direction. Let the north direction be $\

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) The area vector $\vec{A}$ of the loop is perpendicular to the east-west plane,which is in the north-south direction. Let the north direction be $\hat{j}$. Thus,$\vec{A} = (0.1 \ m)^2 \hat{j} = 0.01 \hat{j} \ m^2$.The magnetic field $\vec{B}$ is in the north-east direction,making an angle of $45^\circ$ with the east (horizontal) and north (vertical) axes. Thus,$\vec{B} = 0.20 (\cos 45^\circ \hat{i} + \sin 45^\circ \hat{j}) = 0.20 (\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j}) \ T$.The magnetic flux $\phi$ through the loop is $\phi = \vec{B} \cdot \vec{A} = [0.20 (\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j})] \cdot [0.01 \hat{j}] = 0.20 	imes 0.01 	imes \frac{1}{\sqrt{2}} = \frac{0.002}{\sqrt{2}} = \sqrt{2} 	imes 10^{-3} \ Wb$.The induced emf $e$ is given by $|e| = |\frac{\Delta \phi}{\Delta t}| = |\frac{0 - \phi}{1 \ s}| = \sqrt{2} 	imes 10^{-3} \ V$.Comparing this with $\sqrt{x} 	imes 10^{-3} \ V$,we get $\sqrt{x} = \sqrt{2}$,which implies $x = 2$.

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