Question

a circular coil of $25$ turns and radius $6\text{cm}$ carrying a current of $10\text{A}$, is suspended vertically in a uniform magnetic field of magnitude $1.2\text{T}$. The field lines run horizontally in the plane of the coil. What will be the magnitude and dirction of balancing torque required on the coil to prevent it from turning?

• A. $3.39\text{N-m}$, upwards
• B. $7.5\text{N-m}$, downwards
• C. $3.39\text{N-m}$, downwards
• D. $5\text{N-m}$, upwards

Answer 1

The correct option is C $3.39\text{N-m}$, downwards

The torque $\overrightarrow{\tau }$ on a coil of having $N$ turns and current $i$ placed in a magnetic field will be,

$\overrightarrow{\tau }=\overrightarrow{M}\times \overrightarrow{B}$

$|\overrightarrow{\tau }|=NiAB\sin \theta$

Since, the coil lies in the plane of $\overrightarrow{B}$, $\theta =$ ${90}^{\circ }$

$|\overrightarrow{\tau }|=25\times 10\times [\pi \times (6{)}^2\times {10}^{-4}]\times 1.2\times 1$

$|\overrightarrow{\tau }|=3.39\text{N-m}$

Using Right-hand thumb rule, the direction of magnetic moment $\left(\overrightarrow{M}\right)$ is perpendicularly outwards $(\odot )$, Thus using cross product with horizontal magnetic field (rightwards) give the direction for torque as vertically upwards,


Thus, to prevent the coil from turning an equal & opposite torque need to be applied.

So, the torque must be applied in downward direction.

Hence, option $\left(C\right)$ is the correct answer.