The magnitude of the force per unit length acting on a thin wire carrying a current $I=8 \text{ A}$ at a point $O$, if the wire is bent as shown in the figure with a radius $R=10 \pi \text{ cm}$, is (in $\mu \text{N/m}$)

Two concentric coils each of radius equal to $2\pi \, cm$ are placed at right angles to each other. $3 \, A$ and $4 \, A$ are the currents flowing in each coil respectively. The magnetic induction in $Wb/m^2$ at the centre of the coils will be $(\mu_0 = 4\pi \times 10^{-7} \, Wb/A \cdot m)$.

In the hydrogen atom,the electron is making $6.6 \times 10^{15} \, r.p.s.$ If the radius of the orbit is $0.53 \times 10^{-10} \, m,$ then the magnetic field produced at the centre of the orbit is (in $Tesla$):

If we double the radius of a coil keeping the current through it unchanged,then the magnetic field at any point at a large distance from the centre becomes approximately

The magnitude of the magnetic field at $O$ due to a current-carrying loop as shown in the figure is, where $O$ is the center of two circular portions with radii $1 \, cm$ and $2 \, cm$ respectively. (Take the value of current $I = \frac{1.2}{\pi} \, A$)

$A$ long conducting wire carrying a current $I$ is bent at $120^{\circ}$ (see figure). The magnetic field $B$ at a point $P$ on the angle bisector of the bend at a distance $d$ from the bend is ($\mu_{0}$ is the permeability of free space):