A conducting square loop of side L,mass M and | vedclass

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(B) As the loop enters the magnetic field,the magnetic flux $\phi = B_0 L x$ changes,where $x$ is the distance the loop has entered. The induced $EMF$

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$(B)$ and $(D)$

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(B) As the loop enters the magnetic field,the magnetic flux $\phi = B_0 L x$ changes,where $x$ is the distance the loop has entered. The induced $EMF$ is $\varepsilon = \frac{d\phi}{dt} = B_0 L v$. The induced current is $i = \frac{\varepsilon}{R} = \frac{B_0 L v}{R}$.The magnetic force on the leading edge is $F = i L B_0 = \frac{B_0^2 L^2 v}{R}$. Since this force opposes motion,$M a = -\frac{B_0^2 L^2 v}{R}$.Given $K = \frac{B_0^2 L^2}{RM}$,we have $a = -K v$,or $\frac{dv}{dt} = -K v$.Integrating this,$v(t) = v_0 e^{-Kt}$.The distance $x$ covered is $x(t) = \int_0^t v(t) dt = \frac{v_0}{K}(1 - e^{-Kt})$.For the loop to enter completely,$x$ must reach $L$. Thus $L = \frac{v_0}{K}(1 - e^{-Kt_{entry}})$.$(A)$ If $v_0 = 1.5 KL$,then $L = \frac{1.5 KL}{K}(1 - e^{-Kt}) \Rightarrow 1 = 1.5(1 - e^{-Kt}) \Rightarrow e^{-Kt} = 1 - \frac{1}{1.5} = \frac{1}{3}$. Since $e^{-Kt} > 0$,the loop enters completely. Statement $(A)$ is incorrect.$(B)$ When the loop is fully inside,the flux $\phi = B_0 L^2$ is constant,so $\frac{d\phi}{dt} = 0$,$\varepsilon = 0$,$i = 0$,and $F = 0$. Statement $(B)$ is correct.$(C)$ The loop only comes to rest as $t 	o \infty$. Statement $(C)$ is incorrect.$(D)$ For $v_0 = 3 KL$,$L = \frac{3 KL}{K}(1 - e^{-Kt}) \Rightarrow \frac{1}{3} = 1 - e^{-Kt} \Rightarrow e^{-Kt} = \frac{2}{3} \Rightarrow t = \frac{1}{K} \ln(\frac{3}{2})$. Statement $(D)$ is correct.Thus,the correct statements are $(B)$ and $(D)$.

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