A square frame with side length $l$ has total resistance $R$ . It is being pulled with speed $v$ out of region of uniform magnetic field $B$ . Moment when left corner is at distance $x$ inside field, the force required to pull frame at constant speed $v$ is

\frac{4 B^{2} x^{2} v}{R}

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\frac{2 B^{2} x^{2} v}{2 R}

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\frac{3 B^{2} x^{2} v}{2 R}

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\frac{2 B^{2} x^{2} v}{R}

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Solution

The correct option is A $\frac{4 B^{2} x^{2} v}{R}$

$ε = B v l_{A B}$
$i = \frac{ε}{R} = \frac{B v l_{A B}}{R}$
$F = i l_{A B} B = \frac{B^{2} v l_{A B}^{2}}{R}$
As, $x = \frac{l_{A B}}{2}$

We get,
$F = \frac{B^{2} v 4 x^{2}}{R}$

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A square frame with side length l has total resistance R. It is being pulled with speed v out of region of uniform magnetic field B. Moment when left corner is at distance x inside field, the force required to pull frame at constant speed v is

The correct option is A 4B2x2vR ε=B v lAB i=εR=B v lABR F=i lAB B=B2 v l2ABR As, x=lAB2 We get, F=B2 v 4x2R

A square frame with side length $l$ has total resistance $R$ . It is being pulled with speed $v$ out of region of uniform magnetic field $B$ . Moment when left corner is at distance $x$ inside field, the force required to pull frame at constant speed $v$ is

The correct option is A $\frac{4 B^{2} x^{2} v}{R}$

$ε = B v l_{A B}$
$i = \frac{ε}{R} = \frac{B v l_{A B}}{R}$
$F = i l_{A B} B = \frac{B^{2} v l_{A B}^{2}}{R}$
As, $x = \frac{l_{A B}}{2}$

We get,
$F = \frac{B^{2} v 4 x^{2}}{R}$