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An electron with kinetic energy $5 \mathrm{eV}$ enters a region of uniform magnetic field of $3 \mu \mathrm{T}$ perpendicular to its direction. An electric field $\mathrm{E}$ is applied perpendicular to the direction of velocity and magnetic field. The value of $\mathrm{E}$, so that electron moves along the same path, is . . . . . $\mathrm{NC}^{-1}$.
(Given, mass of electron $=9 \times 10^{-31} \mathrm{~kg}$, electric charge $=1.6 \times 10^{-19} \mathrm{C}$ )
An electron moves along vertical line and away from the observer, then pattern of concentric circular magnetic field lines which are produced due to its motion
A closely wound solenoid of $2000$ $turns$ and area of cross-section $1.5 \times 10^{-4}\ m^2$ carries a current of $2.0\, A.$ It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field $5 \times 10^{- 2}$ $tesla$ making an angle of $30^o $ with the axis of the solenoid. The torque on the solenoid will be
In the given diagram a rod is rotating with angular velocity $\omega $. Mass of this rod is $m$ charge $q$ and length $l$ then find out magnetic moment of this rod
A moving coil galvanometer has a resistance of $50\,\Omega $ and gives full scale deflection for $10\, mA$. How could it be converted into an ammeter with a full scale deflection for $1\,A$
To convert galvanometer into ammeter, shunt of $0.01\,\Omega $ is used. Resistance of galvanometer coil is $50\,\Omega $ and its maximum deflection current is $20\ mA$ . Range of ammeter is
A circular coil of radius $4\, cm$ has $50$ $turns$. In this coil a current of $2\, A$ is flowing. It is placed in a magnetic field of $0.1$ $weber/{m^2}$. The amount of work done in rotating it through $180^\circ $ from its equilibrium position will be........$J$
A single turn current loop in the shape of a right angle triangle with sides $5\,cm , 12\,cm , 13\,cm$ is carrying a current of $2\,A$. The loop is in a uniform magnetic field of magnitude $0.75\,T$ whose direction is parallel to the current in the $13\,cm$ side of the loop. The magnitude of the magnetic force on the $5\,cm$ side will be $\frac{ x }{130}\,N$. The value of $x$ is $..........$