In the figure,the switch $S$ is closed so that a current flows in the iron-core inductor which has inductance $L$ and the resistance $R$. When the switch is opened,a spark is obtained at the contacts. The spark is due to

  • A
    a slow flux change in $L$
  • B
    a sudden increase in the emf of the battery $B$
  • C
    a rapid flux change in $L$
  • D
    a rapid flux change in $R$

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The figure shows a part of a complete circuit. The potential difference $V_B - V_A$ when the current $I$ is $5 \ A$ and is decreasing at a rate of $10^3 \ A \ s^{-1}$ is given by ........ $V$.

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$A$ 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:
$(A)$ If $v_0=1.5 KL$,the loop will stop before it enters completely inside the region of magnetic field.
$(B)$ When the complete loop is inside the region of magnetic field,the net force acting on the loop is zero.
$(C)$ 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 $t=\left(\frac{1}{K}\right) \ln \left(\frac{3}{2}\right)$.

Plane figures made of thin wires of resistance $R = 50 \text{ m}\Omega/\text{m}$ are located in a uniform magnetic field perpendicular to the plane of the figures, which decreases at the rate $dB/dt = 0.1 \text{ mT/s}$. Find the currents in the inner and outer boundary. (The inner radius $a = 10 \text{ cm}$ and outer radius $b = 20 \text{ cm}$)

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