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$(a)$ As wire loop enters the region of magnetic field, an emf is induced in the wire loop. The current of induced emf causes an opposing force on the

Vedclass · Accepted Answer

$v_{0}-\frac{B^{2} a^{2}}{R m} x$

Vedclass · Answer

$(a)$ As wire loop enters the region of magnetic field, an emf is induced in the wire loop. The current of induced emf causes an opposing force on the wire loop.

This force is given by

$F=-BIa=-B\left(\frac{E}{R}\right) \cdot a$

$=-B\left(\frac{B a v}{R}\right) a=\frac{-B^{2} a^{2} v}{R}$

(Negative sign shows retarding force)

So, deacceleration $A$ of loop is

$A =\frac{F}{m}=\frac{-B^{2} a^{2} v}{m R}$

$\Rightarrow \quad \frac{d v}{d t} =\frac{-B^{2} a^{2}}{m R} v \Rightarrow \frac{d v}{d x} \cdot \frac{d x}{d t}=\frac{-B^{2} a^{2} v}{m R}$

$\Rightarrow v \frac{d v}{d x}=\frac{-B^{2} a^{2}}{m R} \cdot v$

$\Rightarrow d v=\frac{-B^{2} a^{2}}{m R} \cdot d x$

Integrating between limits, we have

$\int \limits_{v_{0}}^{v} d v=\int \limits_{0}^{x}-\frac{B^{2} a^{2}}{m R} \cdot d x$

$v-v_{0}=\frac{-B^{2} a^{2}}{m R} \cdot x$

So, velocity of loop at distance $x$ is

$\Rightarrow \quad v=v_{0}-\frac{B^{2} a^{2}}{m R} \cdot x$

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