An $\alpha$ particle is moving along a circle of radius $R$ with a constant angular velocity $\omega$. Point $A$ lies in the same plane at a distance $2R$ from the centre. Point $A$ records the magnetic field produced by the $\alpha$ particle. If the minimum time interval between two successive times at which $A$ records zero magnetic field is $t$,find the angular speed $\omega$ in terms of $t$.

An electron $(q = 1.6 \times 10^{-19}\, C)$ is moving at a right angle to a uniform magnetic field of $3.534 \times 10^{-5}\, T$. The time taken by the electron to complete one circular orbit is ...... $\mu s$.

An electron is moving on a circular path of radius $r$ with speed $v$ in a transverse magnetic field $B$. The $e/m$ ratio for it will be:

If a proton of kinetic energy $8.35 \text{ MeV}$ enters a uniform magnetic field of $10 \text{ T}$ at right angles to the direction of the field,then the force acting on the proton is (Mass of proton $= 1.67 \times 10^{-27} \text{ kg}$ and charge of proton $= 1.6 \times 10^{-19} \text{ C}$)

In the $xy$-plane, the region $y > 0$ has a uniform magnetic field $B_1 \hat{k}$ and the region $y < 0$ has another uniform magnetic field $B_2 \hat{k}$. A positively charged particle is projected from the origin along the positive $y$-axis with speed $v_0 = \pi \text{ m s}^{-1}$ at $t = 0$, as shown in the figure. Neglect gravity in this problem. Let $t = T$ be the time when the particle crosses the $x$-axis from below for the first time. If $B_2 = 4 B_1$, the average speed of the particle, in $\text{m s}^{-1}$, along the $x$-axis in the time interval $T$ is. . . .

An electron with energy $880 \,eV$ enters a uniform magnetic field of induction $2.5 \times 10^{-3} \,T$. The radius of the circular path will approximately be: