Question

A wire $ABC$ carrying current of magnitude $5A$ is placed in a region having horizontal magnetic field $0.15T$ as shown. If the wire $ABC$ resembles a right angled triangle at $A$,length of section $AB$ and $BC$ are $16cm$ and $20cm$ respectively.Then force on the section $BC$ due to magnetic field is____( Assume that current is flowing from $B$ to $C$)

• A. Zero
• B. $0.16N$ outwards
• C. $0.12N$ outwards
• D. $0.12N$ inwards

Answer 1

The correct option is C

$0.12N$ outwards

Force on section $BC$ $=BILsin\theta$
Where $\theta$ is the angle between the section $AB$ and section $BC$.
$\Rightarrow 0.15\times 5\times \frac{20}{100}sin({90}^{\circ }-\theta )=0.15\times 5\times \frac{20}{100}cos\theta =0.15\times 5\times \frac{16}{100}N$
$\Rightarrow 0.12N$
Applying Flemings left hand rule direction of force is outwards.