$A$ square loop of side $5 \ cm$ enters a magnetic field with a velocity of $1 \ cm/s$. The front edge enters the magnetic field at $t = 0$. Which graph best depicts the induced $emf$?

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    Option A
  • B
    Option B
  • C
    Option C
  • D
    Option D

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The bob of a simple pendulum is replaced by a magnet. The oscillations are set along the length of the magnet. $A$ copper coil is added so that one pole of the magnet passes in and out of the coil. The coil is short-circuited. Then which one of the following happens?

$A$ conducting square loop of side $L$,mass $M$ and resistance $R$ is moving in the $XY$ plane with its edges parallel to the $X$ and $Y$ axes. The region $y \geq 0$ has a uniform magnetic field,$\vec{B}=B_0 \hat{k}$. The magnetic field is zero everywhere else. At time $t=0$,the loop starts to enter the magnetic field with an initial velocity $v_0 \hat{\imath} \text{ m/s}$,as shown in the figure. Considering the quantity $K=\frac{B_0^2 L^2}{RM}$ in appropriate units,ignoring self-inductance of the loop and gravity,which of the following statements is/are correct:
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$(C)$ If $v_0=\frac{KL}{10}$,the loop comes to rest at $t=\left(\frac{1}{K}\right) \ln \left(\frac{5}{2}\right)$.
$(D)$ If $v_0=3 KL$,the complete loop enters inside the region of magnetic field at time $t=\left(\frac{1}{K}\right) \ln \left(\frac{3}{2}\right)$.

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