A simple pendulum with bob of mass m and conducting wire of length L swings under gravity through an angle $$2\theta $$. The earth?s magnetic field component in the direction perpendicular to swing is B. Maximum potential difference induced across the pendulum is [MP PET 2005]

Question

A simple pendulum with bob of mass m and conducting wire of length L swings under gravity through an angle $$2\theta $$. The earth?s magnetic field component in the direction perpendicular to swing is B. Maximum potential difference induced across the pendulum is                           [MP PET 2005]

TuteeHUB · Accepted Answer

$$BL\sin \left( \frac{\theta }{2} \right)(gL)$$

TuteeHUB · Answer

$$2BL\sin \left( \frac{\theta }{2} \right){{(gL)}^{1/2}}$$

TuteeHUB · Answer

$$BL\sin \left( \frac{\theta }{2} \right){{(gL)}^{3/2}}$$

TuteeHUB · Answer

$$BL\sin \left( \frac{\theta }{2} \right){{(gL)}^{2}}$$

1.	A simple pendulum with bob of mass m and conducting wire of length L swings under gravity through an angle \[2\theta \]. The earth?s magnetic field component in the direction perpendicular to swing is B. Maximum potential difference induced across the pendulum is [MP PET 2005]
A.	\[2BL\sin \left( \frac{\theta }{2} \right){{(gL)}^{1/2}}\]
B.	\[BL\sin \left( \frac{\theta }{2} \right)(gL)\]
C.	\[BL\sin \left( \frac{\theta }{2} \right){{(gL)}^{3/2}}\]
D.	\[BL\sin \left( \frac{\theta }{2} \right){{(gL)}^{2}}\]
Answer» B. \[BL\sin \left( \frac{\theta }{2} \right)(gL)\]

A simple pendulum with bob of mass m and conducting wire of length L swings under gravity through an angle \[2\theta \]. The earth?s magnetic field component in the direction perpendicular to swing is B. Maximum potential difference induced across the pendulum is [MP PET 2005]

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