First slide

Motional EMF

Question

As shown in the figure a metal rod makes contact and completes the circuit. The circuit is perpendicular to the magnetic field with $B = 0.5 t e s l a .$ If the resistance is $3 Ω$ , work done by magnetic force to move the rod as indicated with a constant speed of $2 m / s e c$ for 3 sec is.

Easy

A

0 J
B

3.5 J
C

1 J
D

4.735 J

Solution

Induced current in the circuit $i = \frac{B v l}{R}$
Magnetic force acting on the wire $F_{m} = B i l = B (\frac{B v l}{R}) l$
$\Rightarrow F_{m} = \frac{B^{2} v l^{2}}{R} =$ External force needed to move the rod with constant velocity
$D i s p l a c e m e n t = S = v t = 2 \times 3 = 6$
Work done $= \overset{}{\vec{F} . \vec{S}} = 1.67 \times 10^{- 1} \times 6 \approx 1 J$

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