Question

A uniform magnetic field of $1.5\text{T}$ exists in a cylindrical region of radius $10\text{cm}$, its direction being parallel to the axis along west to east. A current carrying wire in the south-north direction passes through this region. The wire intersects the axis and experiences a force of $1.2\text{N}$. If the wire is turned to north-east direction, then the magnitude and direction of force will be:

• A. $1.2\text{N}$, upward
• B. $1.2\sqrt{2}\text{N}$, downward
• C. $1.2\text{N}$, downward
• D. $\frac{1.2}{\sqrt{2}}\text{N}$, downward

Answer 1

The correct option is C $1.2\text{N}$, downward

When the wire is carrying current in South-North direction

$\overrightarrow{L}$ is along $+vey-$ direction.

And, $\overrightarrow{B}$ is along $+vex-$ direction.

Now, using the equation of magnetic force,

$\overrightarrow{F}=I(\overrightarrow{L}\times \overrightarrow{B})$

$\overrightarrow{F}=I\left[\right(2r\hat{j})\times (B\hat{i}\left)\right]$

$\overrightarrow{F}=2BIr\sin {90}^{\circ }(-\hat{k})$

$\therefore F=2BIr$, downwards

Given, $F=1.2\text{N}$ directed vertically downward

$\therefore 2BIr=1.2$

Now, when the wire is oriented along north-east, the cross-sectoral view shows;


$\overrightarrow{L}=\overrightarrow{AB}$

From triangle,

$AB=\frac{2r}{\sin \theta }$

$\therefore L=\frac{2r}{\sin \theta }$

using, $\overrightarrow{F}=I(\overrightarrow{L}\times \overrightarrow{B})$

$F=IBL\sin \theta =IB(\frac{2r}{\sin \theta })\sin \theta$

$F=2BIr$

$\therefore F=2BIr=1.2\text{N}$

Direction will still be downward $(-\hat{k})$. It can be found out using $(\overrightarrow{L}\times \overrightarrow{B})$ or right- hand screw rule.

Therefore, option $\left(C\right)$ is right.