$A$ proton of velocity $v = (3 \hat{i} + 2 \hat{j}) \ m/s$ enters a magnetic field of induction $B = (2 \hat{j} + 3 \hat{k}) \ T$. The acceleration produced in the proton in $m/s^2$ is (Specific charge of proton $= 0.96 \times 10^8 \ C/kg$)

An electron enters with a velocity $\vec{v} = v_0 \hat{i}$ into a cubical region (faces parallel to coordinate planes) in which there are uniform electric and magnetic fields. The orbit of the electron is found to spiral down inside the cube in a plane parallel to the $xy$-plane. Suggest a configuration of fields $\vec{E}$ and $\vec{B}$ that can lead to this.

An electron moving with a velocity $\vec{V_1} = 2\,\hat{i}\,\text{m/s}$ at a point in a magnetic field experiences a force $\vec{F_1} = -2\hat{j}\,\text{N}$. If the electron is moving with a velocity $\vec{V_2} = 2\,\hat{j}\,\text{m/s}$ at the same point,it experiences a force $\vec{F_2} = +2\,\hat{i}\,\text{N}$. The force the electron would experience if it were moving with a velocity $\vec{V_3} = 2\hat{k}\,\text{m/s}$ at the same point is

$A$ charged particle is moving in a uniform magnetic field in a circular path of radius $R$. When the energy of the particle becomes three times the original,the new radius will be

The figure shows three situations when an electron moves with velocity $\vec v$ through a uniform magnetic field $\vec B$. In each case,what is the direction of the magnetic force on the electron?

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. . . .