$A$ conducting bar of length $L$ is free to slide on two parallel conducting rails as shown in the figure. Two resistors $R_{1}$ and $R_{2}$ are connected across the ends of the rails. There is a uniform magnetic field $\vec{B}$ pointing into the page. An external agent pulls the bar to the left at a constant speed $v$. The correct statement about the directions of induced currents $I_{1}$ and $I_{2}$ flowing through $R_{1}$ and $R_{2}$ respectively is:

  • A
    Both $I_{1}$ and $I_{2}$ are in anticlockwise direction.
  • B
    Both $I_{1}$ and $I_{2}$ are in clockwise direction.
  • C
    $I_{1}$ is in clockwise direction and $I_{2}$ is in anticlockwise direction.
  • D
    $I_{1}$ is in anticlockwise direction and $I_{2}$ is in clockwise direction.

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$A$ boat is moving due east in a region where the earth's magnetic field is $5.0 \times 10^{-5} \text{ T}$ and is directed due north and horizontal. The boat carries a vertical aerial $2 \text{ m}$ long. If the speed of the boat is $1.5 \text{ m/s}$,calculate the magnitude of the induced $emf$ in the aerial wire in $mV$.

$A$ metal rod of length $L$ rotates about one end at origin with a uniform angular velocity $\omega$. The magnetic field radially falls off as $B(r) = B_0 e^{-\lambda r}$; $\lambda$ being a positive constant. The emf induced (neglecting the centripetal force on electrons in the rod) is :

$A$ rectangular metallic loop is moving out of a uniform magnetic field region to a field-free region with a constant speed. When the loop is partially inside the magnetic field,the plot of the magnitude of induced $\text{emf} \ (\varepsilon)$ with time $(t)$ is given by:

$A$ rectangular coil of single turn,having area $A$,rotates in a uniform magnetic field $B$ with an angular velocity $\omega$ about an axis perpendicular to the field. If initially the plane of the coil is perpendicular to the field,then the average induced $e.m.f.$ when it has rotated through $90^{\circ}$ is

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$A$ metal wire of length $2500 \ m$ is kept in east-west direction,at a certain height from the ground. If it falls freely on the ground,then the current induced in the wire when its speed is $10 \ m/s$ is (Resistance of wire $= 25 \ \Omega$,$g = 10 \ m/s^2$ and Earth's horizontal component of magnetic field $B_{H} = 2 \times 10^{-5} \ T$). (in $A$)

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