Solution:

Ans.

$$B=\frac{{\mu }_{0}I{r}^2}{2{(r^2+x^2)}^{3/2}}$$

Net field at $O,B_0=\sqrt{2}B=\frac{\sqrt{2}{\mu }_0\mathrm{I}{r}^2}{2{(r^2+x^2)}^{3/2}}1/2$

For small loop $(r\ll x),B_0=\frac{\sqrt{2}{\mu }_0\mathrm{I}}{2x^3}{r}^2$

$1/2$

The direction of $B_0$ is ${45}^{\circ }$ with either of the field.

$1/2$

[CBSE Marking Scheme 2014]

[II Q. 3. Two identical circular wires $P$ and $Q$ each of radius $R$ and carrying current $I$ are kept in perpendiculay planes such that they have a common centre as shown in the figure. Find the magnitude and direction of the net magnetic field at the common centre of the two coils.

A [Delhi I, II, III 2012]

Ans. We have : $B_P=B_Q=\frac{{\mu }_{0}I}{2R}$

$1/2$

$B_P$ is directed in the vertically upward direction while $B_Q$ is directed along the horizontal direction.

$1/2$

$$ \begin{aligned} \therefore \quad B & =\sqrt{B^{2}{ }{P}+B^{2}{ }{Q}} \ & =\sqrt{2} B_{P} \ & =\sqrt{2} B_{Q} \ \Rightarrow \quad B & =\sqrt{2} \frac{\mu_{0} I}{2 R} \ & =\left(\frac{\mu_{0} I}{\sqrt{2} R}\right) \end{aligned} $$

$1/2$

The net magnetic field is directed at an angle of ${45}^{\circ }$ with either of the field.

?. Short Answer Type Questions-II

[AI Q. 1. (a) Derive an expression for the velocity $\overrightarrow{v_c}$ of a positive ion passing undeflected through a region where crossed and uniform electric field $\overrightarrow{E}$ and magnetic field $\overrightarrow{B}$ are simultaneously present. (b) Draw and justify the trajectory of identical positive ion whose velocity has a magnitude less than $\left|{\overrightarrow{v}}_c\right|$.

R&U [SQP 2018]

Ans. (a) $\overrightarrow{E}=E\hat{j}$ and $\overrightarrow{B}=B\hat{k}$

Force on positive ion due to electric field

$${\overrightarrow{F}}_E=qE\hat{j}$$

Force due to magnetic field $\vec{F}{B}=q\left(\overrightarrow{v{c}} \times \vec{B}\right)$

For passing undeflected, $\vec{F}{E}=-\overrightarrow{F{B}}$

$$qE\hat{j}=-q(\overrightarrow{v_c}\times B\hat{k})$$

This is possible only if $q \vec{v}{c} \times B \hat{k}=q v{c} B \hat{j}$ or

$${\overrightarrow{v}}_c=(E/B)\hat{i}$$

(b) The trajectory would be as shown.

Justification: For positive ions with speed $v<v_c$

Force due to electric field $=F_E^{\prime }=qE=F_E$ due to magnetic field $F_B^{\prime }=qvB<F_B$ since $v<v_c$ Now forces are unbalanced, and hence, ion will experience an acceleration along $E$. Since initial velocity is perpendicular to $E$, the trajectory would be parabolic.

[CBSE Marking Scheme 2018]