A conducting rod PQ of length L = 1.0 m is moving with a uniform speed v = 2 m/s in a uniform magnetic field B = 4.0 T directed into the paper. A capacitor of capacity C = 10 μF is connected as shown in figure. Then

1. q_A = + 80 μC and q_B = – 80 μC

2. q_A = – 80 μC and q_B = + 80 μC

3. q_A = 0 = q_B

4. Charge stored in the capacitor increases exponentially with time

The resistance in the following circuit is increased at a particular instant. At this instant the value of resistance is \(10~\Omega.\) The current in the circuit will be:

Shown in the figure is a circular loop of radius r and resistance R. A variable magnetic field of induction B = B₀e^–t is established inside the coil. If the key (K) is closed, the electrical power developed right after closing the switch, at t=0, is equal to

1. $\frac{B_0^2\pi r^2}{R}$

2. $\frac{B_{0}10r^3}{R}$

3. $\frac{B_0^2{\pi }^2{r}^{4}R}{5}$

4. $\frac{B_0^2{\pi }^2{r}^4}{R}$

A highly conducting ring of radius R is perpendicular to and concentric with the axis of a long solenoid as shown in fig. The ring has a narrow gap of width d in its circumference. The solenoid has a cross-sectional area A and a uniform internal field of magnitude B₀. Now beginning at t = 0, the solenoid current is steadily increased so that the field magnitude at any time t is given by B(t) = B₀ + αt where α > 0. Assuming that no charge can flow across the gap, the end of the ring which has an excess of positive charge and the magnitude of induced e.m.f. in the ring are respectively

1. X, Aα

2. X πR²α

3. Y, πA²α

4. Y, πR²α

A rectangular loop with a sliding connector of length \(l= 1.0\) m is situated in a uniform magnetic field \(B = 2T\) perpendicular to the plane of the loop. Resistance of connector is \(r=2~\Omega\). Two resistances of \(6~\Omega\) and \(3~\Omega\) are connected as shown in the figure. The external force required to keep the connector moving with a constant velocity \(v = 2\) m/s is:

          
1. \(6~\text{N}\)
2. \(4~\text{N}\)
3. \(2~\text{N}\)
4. \(1~\text{N}\)

A wire cd of length l and mass m is sliding without friction on conducting rails ax and by as shown. The vertical rails are connected to each other with a resistance R between a and b. A uniform magnetic field B is applied perpendicular to the plane abcd such that cd moves with a constant velocity of

1. $\frac{mgR}{Bl}$

2. $\frac{mgR}{B^2{l}^2}$

3. $\frac{mgR}{B^3{l}^3}$

4. $\frac{mgR}{B^{2}l}$

A conducting rod AC of length 4l is rotated about a point O in a uniform magnetic field $\overrightarrow{B}$ directed into the paper. AO = l and OC = 3l. Then

1. $V_A-V_O=\frac{B\omega l^2}{2}$

2. $V_O-V_C=\frac{7}{2}B\omega l^2$

3. $V_A-V_C=4B\omega l^2$

4. $V_C-V_O=\frac{9}{2}B\omega l^2$

The figure shows three circuits with identical batteries, inductors, and resistors. Rank the circuits according to the current, in descending order, through the battery \((i)\) just after the switch is closed and \((ii)\) a long time later:

The network shown in the figure is a part of a complete circuit. If at a certain instant the current i is 5 A and is decreasing at the rate of 10³ A/s then V_B – V_A is

1. 5 V

2. 10 V

3. 15 V

4. 20 V

A simple pendulum with bob of mass m and conducting wire of length L swings under gravity through an angle 2θ. The earth’s magnetic field component in the direction perpendicular to swing is B. Maximum potential difference induced across the pendulum is 

1. $2BL\sin \left(\frac{{\displaystyle \theta }}{{\displaystyle 2}}\right){\left(gL\right)}^{1/2}$

2. $BL\sin \left(\frac{{\displaystyle \theta }}{{\displaystyle 2}}\right){\left(gL\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

3. $BL\sin \left(\frac{{\displaystyle \theta }}{{\displaystyle 2}}\right){\left(gL\right)}^{3/2}$

4. $BL\sin \left(\frac{{\displaystyle \theta }}{{\displaystyle 2}}\right){\left(gL\right)}^2$