A long solenoid of radius $2\, cm$ has $100\, turns/cm$ and carries a current of $5\,A$. A coil of radius $1\, cm$ having $100\, turns$ and a total resistance of $20\,\Omega $ is placed inside the solenoid coaxially. The coil is connected to a galvanometer. If the current in the solenoid is reversed in direction, find the charge flown through the galvanometer
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
A tightly wound $100$ turns coil of radius $10 \mathrm{~cm}$ carries a current of $7 \mathrm{~A}$. The magnitude of the magnetic field at the centre of the coil is (Take permeability of free space as $4 \pi \times 10^{-7} \mathrm{SI}$ units):
A proton and an $\alpha - $particle enter a uniform magnetic field perpendicularly with the same speed. If proton takes $25$ $\mu \, sec$ to make $5$ revolutions, then the periodic time for the $\alpha - $ particle would be........$\mu \, sec$
Magnitude of magnetic field (in $SI$ units) at the centre of a hexagonal shape coil of side $10\, cm$, $50$ turns and carrying current $I$ (Ampere) in units of $\frac{\mu_{0} I}{\pi}$ is
A non conducting ring (of mass $m,$ radius $r,$ having charge $Q$) is placed on a rough horizontal surface (in a region with transverse magnetic field). The field is increasing with time at the rate $R$ and coefficient of friction between the surface and the ring is $\mu .$ For ring to remain in equilibrium $\mu$ should be greater than
A rectangular coil of length $0 .12\, m$ and width $0.1\, m$ having $50$ turns of wire is suspended vertically in a uniform magnetic field of strength $0.2\,\, Weber/m^2.$ The coil carries a current of $2 \,\,A.$ If the plane of the coil is inclined at an angle of $30^o $ with the direction of the field, the torque required to keep the coil in stable equilibrium will be .......$Nm$
A proton is projected with a velocity $10^7\, m/s$, at right angles to a uniform magnetic field of induction $100\, mT$. The time (in second) taken by the proton to traverse $90^o$ arc is $(m_p = 1.65\times10^{-27}\, kg$ and $q_p = 1.6\times10^{-19}\, C)$