Solution:

Ans.

$$ B=\frac{\mu_{0} I r^{2}}{2\left(r^{2}+x^{2}\right)^{3 / 2}} $$

Net field at $O, \quad B_{0}=\sqrt{2} B=\frac{\sqrt{2} \mu_{0} \mathrm{I} r^{2}}{2\left(r^{2}+x^{2}\right)^{3 / 2}} \quad 1 / 2$

For small loop $(r \ll x), B_{0}=\frac{\sqrt{2} \mu_{0} \mathrm{I}}{2 x^{3}} r^{2}$

$1 / 2$

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 : $B_{P}=B_{Q}=\frac{\mu_{0} I}{2 R}$

$1 / 2$

$B_{P}$ 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 $\vec{E}$ and magnetic field $\vec{B}$ are simultaneously present. (b) Draw and justify the trajectory of identical positive ion whose velocity has a magnitude less than $\left|\vec{v}_{c}\right|$.

R&U [SQP 2018]

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

Force on positive ion due to electric field

$$ \vec{F}_{E}=q E \hat{j} $$

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

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

$$ q E \hat{j}=-q\left(\overrightarrow{v_{c}} \times B \hat{k}\right) $$

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

$$ \vec{v}_{c}=(E / B) \hat{i} $$

(b) The trajectory would be as shown.

Justification: For positive ions with speed $v<v_{c}$

Force due to electric field $=F_{E}^{\prime}=q E=F_{E}$ due to magnetic field $F_{B}^{\prime}=q v B<F_{B}$ 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]