A particle moving with velocity $v$ having specific charge $\left(q/m\right)$ enters a region of magnetic field $B$ having width $d=\frac{3mv}{5qB}$ at angle ${53}^{\circ }$ to the boundary of magnetic field. Find the angle $\theta$ in the diagram.

 

(1) The magnetic field strength may have been increased while the particle was travelling in air
(2) The particle lost energy by ionising the air
(3) The particle lost charge by ionising the air

A negative charge is given to a nonconducting loop and the loop is rotated in the plane of paper about its centre as shown in figure. The magnetic field produced by the ring affects a small magnet placed above the ring in the same plane :

A current element $∆\overrightarrow{l}=∆x\hat{i}-∆y\hat{j}$ carries $10\mathrm{A}$ current. It is placed at origin. Magnitude of magnetic field at point '$P$' which is at position vector $\overrightarrow{r}=\left(\hat{i}+\hat{j}\right)m$ with respect to origin is $B$ (where $∆x=∆y=1\mathrm{mm}$). 

Find $B\times {10}^{10}$ (In SI units).

A pair of stationary and infinitely long bent wires are placed in the $x-y$ plane as shown in the figure. The wires carry currents of $i=10 \mathrm{A}$ each as shown. The segments $L$ and $M$ are along the $x$-axis. The segment $P$ and $Q$ are parallel to the $y$-axis such that $OS=OR=0.02 \mathrm{m}$. Final value of, $B\times {10}^4$, where $B$ is the magnitude of the magnetic induction at the origin $O$ (in SI units)

Find the magnetic field at the centre $P$ of square of side $a=2\mathrm{mm}$ shown in figure (in SI unit). $\left(i={10}^4\mathrm{A}\right)$

A rectangular loop of side $a=4\mathrm{m}$ and $b=3\mathrm{m}$ is carrying the current $I=\frac{3}{10}\mathrm{A}$, then find out value of $B\times {10}^7$, where $B$ is the magnetic field (in SI unit) at the centre of loop.

If magnetic field at point $O$ is zero then find out the value of $\theta$ (in radian).