A simple pendulum with a bob of mass m and a | vedclass

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(A) The motional e.m.f. induced in a conductor of length $L$ moving with velocity $v$ in a magnetic field $B$ is given by $\varepsilon = BLv$. For a p

Vedclass · Accepted Answer

$2 BL(\sqrt{gL})(\sin 	heta / 2)$

Vedclass · Answer

(A) The motional e.m.f. induced in a conductor of length $L$ moving with velocity $v$ in a magnetic field $B$ is given by $\varepsilon = BLv$. For a pendulum,the maximum velocity $v_{max}$ occurs at the lowest point of the swing. Using the principle of conservation of energy,the potential energy at the maximum angle $	heta$ is converted into kinetic energy at the lowest point: $mgL(1 - \cos 	heta) = \frac{1}{2}mv_{max}^2$. Simplifying this,$v_{max}^2 = 2gL(1 - \cos 	heta)$. Using the trigonometric identity $1 - \cos 	heta = 2 \sin^2(	heta / 2)$,we get $v_{max}^2 = 2gL(2 \sin^2(	heta / 2)) = 4gL \sin^2(	heta / 2)$. Taking the square root,$v_{max} = 2\sqrt{gL} \sin(	heta / 2)$. Substituting $v_{max}$ into the e.m.f. formula: $\varepsilon_{max} = BL(2\sqrt{gL} \sin(	heta / 2)) = 2BL\sqrt{gL} \sin(	heta / 2)$. Thus,the correct option is $A$.

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