Moving perpendicular to field $B$, a proton and an alpha particle both enter an area of uniform magnetic field $B$. If the kinetic energy of the proton is $1~\text{MeV}$ and the radius of the circular orbits for both particles is equal, the energy of the alpha particle will be:
1. $4~\text{MeV}$
2. $0.5~\text{MeV}$
3. $1.5~\text{MeV}$
4. $1~\text{MeV}$

A rectangular coil of length $0.12~\text{m}$ and width $0.1~\text{m}$ having $50$ turns of wire is suspended vertically in a uniform magnetic field of strength $0.2~\text{Wb/m}^2$. The coil carries a current of $2~\text{A}$. If the plane of the coil is inclined at an angle of $30^{\circ}$ with the direction of the field, the torque required to keep the coil in stable equilibrium will be:
1. $0.15~\text{N-m}$
2. $0.20~\text{N-m}$
3. $0.24~\text{N-m}$
4. $0.12~\text{N-m}$

A wire carrying current $I$ has the shape as shown in the adjoining figure. Linear parts of the wire are very long and parallel to $X$-axis while the semicircular portion of radius $R$ is lying in the $Y\text-Z$ plane. The magnetic field at point $O$ is:

An electron moving in a circular orbit of radius $r$ makes $n$ rotations per second. The magnetic field produced at the centre has a magnitude:
1. $\frac{\mu_0ne}{2\pi r}$
2. zero
3. $\frac{n^2e}{r}$
4. $\frac{\mu_0ne}{2r}$

Two identical long conducting wires $({AOB})$ and $({COD})$ are placed at a right angle to each other, with one above the other such that '$O$' is the common point for the two. The wires carry $I_1$ and $I_2$ currents, respectively. The point '$P$' is lying at a distance '$d$' from '$O$' along a direction perpendicular to the plane containing the wires. What will be the magnetic field at the point $P?$