Question

A current carrying wire $\text{LN}$ is bent in the form shown below. If the wire carries a current of $10\text{A}$ and it is placed in a magnetic field of $5\text{T}$ which acts perpendicular to the plane of paper and points outwards then it will experience a force,

• A. Zero
• B. $5\text{N}$
• C. $30\text{N}$
• D. $20\text{N}$

Answer 1

The correct option is B $5\text{N}$

Upon observing the wire $\text{LN}$ we find that the shape of the wire is symmetrical with respect to the horizontal wire, $\text{AB}$.


The effective length vector $\overrightarrow{LN}$ joining points L and N will be passing through the mid-point of section $\text{AB}$ due to symmetry of shape

$AO=OB=\frac{6}{2}=3\text{cm}$

Thus, $LO=ON=\sqrt{3^2+4^2}=5\text{cm}=5\times {10}^{-2}m$

From the figure angle between ${\overrightarrow{L}}_{AN}$ and $\overrightarrow{B}$ is ${90}^{\circ }$

Using relation:

$|\overrightarrow{F}|=|i({\overrightarrow{L}}_{AN}\times \overrightarrow{B})|=Bi{L}_{AN}\sin {90}^{\circ }$

$\Rightarrow F=Bi{L}_{AN}$

$F=5\times 10\times \left(2L_0\right)=50\times (2\times 5\times {10}^{-2})$

$\therefore F=5\text{N}$