$A$ charged particle enters a magnetic field $B$ with its initial velocity making an angle of $45^\circ$ with $B$. The path of the particle will be

In a region of space,a uniform magnetic field $B$ exists in the $y-$direction. $A$ proton is fired from the origin,with its initial velocity $v$ making a small angle $\alpha$ with the $y-$direction in the $yz$ plane. In the subsequent motion of the proton,

An electron is moving with a speed of $10^8 \ m/s$ perpendicular to a uniform magnetic field of intensity $B$. Suddenly,the intensity of the magnetic field is reduced to $B/2$. The radius of the path becomes from the original value of $r$:

$A$ rod $AB$ moves with a uniform velocity $v$ in a uniform magnetic field as shown in the figure.

$A$ charged particle moves through a magnetic field perpendicular to its direction. Then

An electron moves with speed $2 \times 10^5 \ m/s$ along the positive $x$-direction in the presence of a magnetic field of induction $B = \hat{i} + 4\hat{j} - 3\hat{k} \ T$. The magnitude of the force experienced by the electron in newtons is (Charge on the electron $= 1.6 \times 10^{-19} \ C$)