A conducting bar of mass m and length l moves | vedclass

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(C) As the bar moves to the right,the area of the loop increases,which increases the magnetic flux through the loop. According to Lenz's law,the induc

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distance the bar travels before it comes to a complete stop is proportional to $R$

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(C) As the bar moves to the right,the area of the loop increases,which increases the magnetic flux through the loop. According to Lenz's law,the induced current will flow in a direction to oppose this increase in flux. Thus,the current flows counter-clockwise.The magnetic force on the bar is $F = IlB = (\frac{Blv}{R})lB = \frac{B^2l^2v}{R}$. This force acts to the left,opposing the motion.Using Newton's second law,$ma = -F = -\frac{B^2l^2v}{R}$.Thus,$m \frac{dv}{dt} = -\frac{B^2l^2v}{R}$. This shows that the velocity decreases exponentially with time,not linearly.To find the distance $x$,we use $v \frac{dv}{dx} = -\frac{B^2l^2v}{Rm}$.Integrating $dv = -\frac{B^2l^2}{Rm} dx$ from $v_0$ to $0$ and $0$ to $x_{max}$:$v_0 = \frac{B^2l^2}{Rm} x_{max} \Rightarrow x_{max} = \frac{m v_0 R}{B^2l^2}$.Therefore,the distance traveled is proportional to $R$.

$A$ conducting bar of mass $m$ and length $l$ moves on two frictionless parallel rails in the presence of a constant uniform magnetic field of magnitude $B$ directed into the page as shown in the figure. The bar is given an initial velocity $v_{0}$ towards the right at $t=0$. Then,the:

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