$1.8 \times 10^{-4} \text{ volt}$
Using Faraday’s law,
emf = B × l × v
Given:
B = 0.3 T
l = 3 cm = 0.03 m
v = 2 cm/s = 0.02 m/s
So,
emf = 0.3 × 0.03 × 0.02
= 1.8 × 10-4 V
Therefore, the induced emf is 1.8 × 10-4 volt.
In the circuit shown below, the voltage appearing across the diode D will be of the form :
A 100-turn closely wound circular coil of radius $10 \text{ cm}$ has a magnetic field of $3.14 \times 10^{-3} \text{ T}$ at its centre. The current passing through the coil, and the magnitude of the magnetic moment of this coil are, respectively :
(Take $\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$)
The current $I$ in the circuit shown below is :
(All diodes are ideal and identical.

The figure given below shows a long straight solid wire of circular cross-section of radius 'a' carrying steady current I. The current I is uniformly distributed across its cross-section. The plot which correctly represents the variation of magnetic field (B) with distance (r) from the axis of the conductor in the region is :
Five capacitors of capacitances $C_1 = C_2 = C_3 = C_4 = 10 \ \mu\text{F}$ and $C_5 = 2.5 \ \mu\text{F}$ are connected as shown along with a battery of $50 \text{ V}$. 
The equivalent capacitance and the charges on each capacitor respectively are :
Three long straight wires carrying current are arranged mutually parallel as shown in the figure. The force experienced by $15 \text{ cm}$ length of wire $Q$ is________.
$(\mu_o = 4\pi \times 10^{-7} \text{ T.m/A})$