A simple pendulum made of mass 10 g and a metallic wire of length 10 cm is suspended vertically in a uniform magnetic field of 2 T . The magnetic field direction is perpendicular to the plane of oscillations of the pendulum. If the pendulum is released from an angle of $60^{\circ}$ with vertical, then maximum induced EMF between the point of suspension and point of oscillation is
$\_\_\_\_$ mV . (Take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )
A conducting circular loop is rotated about its diameter at a constant angular speed of $100 \mathrm{rad} / \mathrm{s}$ in a magnetic field of 0.5 T perpendicular to the axis of rotation. When the loop is rotated by $30^{\circ}$ from the horizontal position, the induced EMF is 15.4 mV . The radius of the loop is $\_\_\_\_$ mm.
$$ \left(\text { Take } \pi=\frac{22}{7}\right) $$
Inductance of a coil with $10^4$ turns is 10 mH and it is connected to a dc source of 10 V with internal resistance of $10 \Omega$. The energy density in the inductor when the current reaches $\left(\frac{1}{e}\right)$ of its maximum value is $\alpha \pi \times \frac{1}{e^2} \mathrm{~J} / \mathrm{m}^3$. The value of $\alpha$ is
$\_\_\_\_$ .
$$ \left(\mu_0=4 \pi \times 10^{-7} \mathrm{Tm} / \mathrm{A}\right) . $$
Conductor wire ABCDE with each arm 10 cm in length is placed in magnetic field of $\frac{1}{\sqrt{2}}$ Tesla, perpendicular to its plane. When conductor is pulled towards right with constant velocity of $10 \mathrm{~cm} / \mathrm{s}$, induced emf between points A and E is ________ mV .

JEE Main Subjects
Browse all chapters by subject