Question: Q. 2. Two very small identical circular loops, (1) and (2), carrying equal currents $I$ are placed vertically (with respect to the plane of the paper) with their geometrical axes perpendicular to each other as shown in the figure. Find the magnitude and direction of the net magnetic field produced at the point $O$.

A [Foreign I, II, III 2014]

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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]



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