$(a)$ $A$ closed loop is held stationary in the magnetic field between the north and south poles of two permanent magnets held fixed. Can we hope to generate current in the loop by using very strong magnets?
$(b)$ $A$ closed loop moves normal to the constant electric field between the plates of a large capacitor. Is a current induced in the loop
$\quad (i)$ when it is wholly inside the region between the capacitor plates
$\quad (ii)$ when it is partially outside the plates of the capacitor? The electric field is normal to the plane of the loop.
$(c)$ $A$ rectangular loop and a circular loop are moving out of a uniform magnetic field region to a field-free region with a constant velocity $v$. In which loop do you expect the induced emf to be constant during the passage out of the field region? The field is normal to the loops.
$(d)$ Predict the polarity of the capacitor in the situation described by the figure.

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(N/A) No. Current can be induced in a closed loop only when the magnetic flux linked with the loop changes with time. Since the loop and the magnets are stationary,the magnetic flux remains constant,and hence no current is generated,regardless of the strength of the magnets.
$(b)$ No current is induced in either case. Electromagnetic induction is caused by a change in magnetic flux,not electric flux. Since the electric field is constant and the loop moves normal to it,the electric flux through the loop may change,but this does not induce an electric current in the loop.
$(c)$ The induced emf is given by $\varepsilon = Blv$,where $l$ is the length of the side of the loop cutting the magnetic field lines. For a rectangular loop,the length $l$ perpendicular to the velocity $v$ remains constant as it exits the field,so the induced emf is constant. For a circular loop,the length of the chord cutting the field lines changes continuously as it exits,so the induced emf varies.
$(d)$ According to Lenz's Law,the induced current will oppose the cause of its production. As the magnets move towards the loop,the magnetic flux through the loop increases. The induced current will create a magnetic field to oppose this increase. Using the right-hand rule,the induced current will flow in a direction such that plate $A$ becomes positive with respect to plate $B$.

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$A$ square loop of side $1 \, m$ is placed in a perpendicular magnetic field. Half of the area of the loop is inside the magnetic field. $A$ battery of $emf$ $10 \, V$ and negligible internal resistance is connected in the loop. The magnetic field changes with time according to the relation $B = (0.01 - 2t) \, Tesla$. The resultant $emf$ in the loop will be.....$V$

$A$ square loop of side $0.1 \, m$ and resistance $1 \, \Omega$ is moved with a constant velocity in a magnetic field of $2 \, Wb/m^2$. If a current of $1 \, mA$ is induced in the circuit, what is the velocity of the loop in $cm/sec$? (The external circuit consists of a Wheatstone bridge with five $3 \, \Omega$ resistors as shown in the figure.)

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List-$I$List-$II$
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$(Q)$ At $t = 0.2 \text{ s}$,the magnitude of the magnetic force in Newton$(2)$ $0.144$
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$(5)$ $2.00$

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The switch $S$ shown in the circuit is closed at $t=0$. The ratio of the current drawn from the battery by the circuit at $t=0$ and $t=\infty$ is:

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