First slide

Magnetic flux and induced emf

Question

A conducting wire of mass m slides down two smooth conducting bars, set at an angle $θ$ to the horizontal as shown in figure. The separation between the bars is l. The
system is located in the magnetic field B, perpendicular to the plane of the sliding wire and bars. The constant velocity of the wire is

Moderate

A

$\frac{m g R \sin θ}{B^{2} l^{2}}$
B

$\frac{m g R \sin θ}{B l^{3}}$
C

$\frac{m g R θ}{B^{2} l^{5}}$
D

$\frac{m g R \sin θ}{B l^{4}}$

Solution

Component of weight along the inclined plane = mg sin $θ$
$Again, F_{m} = B i l = B \frac{B l v}{R} l = \frac{B^{2} l^{2} v}{R}$
For constant velocity, the magnetic force is balanced by component of weight along the inclined plane.
$Now, \frac{B^{2} l^{2} v}{R} = m g \sin θ or v = \frac{m g R \sin θ}{B^{2} l^{2}}$

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