A circular loop of radius 1m is kept in a magnetic field of strength 2 T directed perpendicular to the plane of loop. Resistance of the loop wire is 2 m. A conductor of length 2 m is sliding with a speed 1 m/s as shown in the figure. Find the instantaneous force acting on the rod [Assume that the rod has negligible resistance]

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Solution

The correct option is B $16$ N
Let induced emf is (e) = V B l
where V = velocity of conductor , B = magnetic field intensity , l = length of conductor.
Current is I.If resistance of loop is 2 R then circuit will be .
current $I = \frac{e}{R_{1} ∥ R_{2}}$
$= \frac{2 e}{1} = 2 V B l$
Force $F_{m} = i l B = 2 v B^{2} l^{2} = 2 \times 1 \times 2^{2} \times 2 F_{m} = 16 N$

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