First slide

Magnetic flux and induced emf

Question

A conducting rod of length $l$ is hinged at point O. It is free to rotate in a vertical plane. There exists a uniform magnetic field $\vec{B}$ in horizontal direction. The rod is released from the horizontal position as shown. The potential difference between the ends of the rod when it has turned by $θ$ (as shown in figure) is proportional to

Moderate

A

$l^{1 / 2}$
B

$l^{2}$
C

$\sin θ$
D

${(\sin θ)}^{1 / 2}$

Solution

The induced electro motive force $(Δ v)$ across ends of rotating rod about one end in magnetic field $= \frac{B}{2} l^{2} ω$ .
$ω$ can be calculated according to law of conservation energy.
Increase in rotational kinetic energy = decrease in gravitational potential energy
$\frac{1}{2} I ω^{2} = m g h$
$\Rightarrow \frac{1}{2} \frac{m l^{2}}{3} ω^{2} = m g . \frac{l}{2} \sin θ$
$\Rightarrow ω = \sqrt{\frac{3 g}{l} \sin θ}$
Now $Δ V = \frac{1}{2} B l^{2} \sqrt{\frac{3 g}{l} \sin θ}$ ……( 1 )
$∴ Δ V \propto \sqrt{\sin θ}$
Also from ( 1 ), $Δ V \propto l^{\frac{3}{2}}$
Therefore, the correct answer is ( D )

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