Two parallel long straight conductors lie on a smooth surface. Two other parallel conductors rest on them at right angles so as to form a square of side $a$ initially. $A$ uniform magnetic field $B$ exists at right angles to the plane containing the conductors. They all start moving outwards with a constant velocity $v$. If $r$ is the resistance per unit length of the wire,the current in the circuit will be

$A$ metallic disc of radius $0.3 \ m$ is rotating with a constant angular speed of $60 \ rad \ s^{-1}$ in a plane perpendicular to a uniform magnetic field of $5 \times 10^{-2} \ T$. The emf induced between a point on the rim and the centre of the disc is: (in $V$)

$A$ coil of $N$ turns and mean cross-sectional area $A$ is rotating with uniform angular velocity $\omega$ about an axis at right angle to a uniform magnetic field $B$. The induced e.m.f. $E$ in the coil will be

$A$ conducting wire $XY$ of mass $m$ and negligible resistance slides smoothly on two parallel conducting wires as shown in the figure. The closed circuit has a resistance $R$ due to $AC$. $AB$ and $CD$ are perfect conductors. There is a magnetic field $\vec{B} = B(t) \hat{k}$.
$(i)$ Write down the equation for the acceleration of the wire $XY$.
$(ii)$ If $\vec{B}$ is independent of time,obtain $v(t)$,assuming $v(0) = u_0$.
$(iii)$ For $(ii)$,show that the decrease in kinetic energy of $XY$ equals the heat lost in $R$.

$A$ circular coil of radius $10\; cm$,$500$ turns,and resistance $2\; \Omega$ is placed with its plane perpendicular to the horizontal component of the Earth's magnetic field. It is rotated about its vertical diameter through $180^{\circ}$ in $0.25\; s$. Estimate the magnitudes of the emf and current induced in the coil. The horizontal component of the Earth's magnetic field at the place is $3.0 \times 10^{-5}\; T$.

The figure shows a straight wire placed between the pole pieces of a magnet. An induced emf will be developed across the ends of the wire when it is moved towards: