A square-shaped wire loop of mass m,resistanc | vedclass

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(A) As the wire loop enters the region of the magnetic field,an electromotive force (emf) is induced in the loop. The current due to this induced emf

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

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(A) As the wire loop enters the region of the magnetic field,an electromotive force (emf) is induced in the loop. The current due to this induced emf causes an opposing magnetic force on the wire loop.The induced emf is given by $E = B a v$.The induced current is $I = \frac{E}{R} = \frac{B a v}{R}$.The magnetic force on the loop is $F = -B I a = -B \left( \frac{B a v}{R} ight) a = -\frac{B^{2} a^{2} v}{R}$.(The negative sign indicates that the force is a retarding force).Using Newton's second law,the acceleration $A$ of the loop is:$A = \frac{F}{m} = \frac{d v}{d t} = -\frac{B^{2} a^{2} v}{m R}$.We can rewrite this using the chain rule $\frac{d v}{d t} = \frac{d v}{d x} \cdot \frac{d x}{d t} = v \frac{d v}{d x}$:$v \frac{d v}{d x} = -\frac{B^{2} a^{2} v}{m R}$.Dividing both sides by $v$ (assuming $v 
eq 0$):$d v = -\frac{B^{2} a^{2}}{m R} d x$.Integrating both sides with limits from $v_{0}$ to $v$ and $0$ to $x$:$\int_{v_{0}}^{v} d v = -\frac{B^{2} a^{2}}{m R} \int_{0}^{x} d x$.$v - v_{0} = -\frac{B^{2} a^{2}}{m R} x$.Therefore,the velocity of the loop at distance $x$ is:$v = v_{0} - \frac{B^{2} a^{2}}{m R} x$.

$A$ square-shaped wire loop of mass $m$,resistance $R$,and side $a$ moving with speed $v_{0}$,parallel to the $X$-axis,enters a region of uniform magnetic field $B$,which is perpendicular to the plane of the loop. The speed of the loop changes with distance $x$ $(x < a)$ in the field as:

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