Two straight conducting plates form an angle | vedclass

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(C) Let the distance of the bar from the vertex at time $t$ be $x = vt$.In the isosceles triangle formed by the plates and the bar, the length of the

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$2 Bv^2 	an \frac{	heta}{2}$

Vedclass · Answer

(C) Let the distance of the bar from the vertex at time $t$ be $x = vt$.In the isosceles triangle formed by the plates and the bar, the length of the bar $\ell$ can be calculated using trigonometry.Considering the right-angled triangle formed by the bisector of the angle $	heta$, we have:$\frac{\ell/2}{x} = 	an \frac{	heta}{2}$$\ell = 2x 	an \frac{	heta}{2} = 2vt 	an \frac{	heta}{2}$.The motional emf induced in a conductor of length $\ell$ moving with velocity $v$ in a magnetic field $B$ is given by $\varepsilon = B \ell v$.Substituting the value of $\ell$ at $t = 1 	ext{ s}$:$\ell = 2v(1) 	an \frac{	heta}{2} = 2v 	an \frac{	heta}{2}$.Therefore, the induced emf is:$\varepsilon = B(2v 	an \frac{	heta}{2})v = 2Bv^2 	an \frac{	heta}{2}$.

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