Question

A rectangular coil of effective area \(0.04\text{ m}^{2}\) is suspended freely in a radial magnetic field of \(0.02\text{ Wb/m}^{2}\). The torsional constant of the suspension fibre is \(5\times 10^{-9}\text{ Nm/degree}\). If a current of \(250\ \mu \text{A}\) is passed through it, then the angle through which it rotates is

• A. \(30^{\circ }\)
• B. \(40^{\circ }\)
• C. \(45^{\circ }\)
• D. \(50^{\circ }\)

Hint:

The magnetic torque NIAB balances the restoring torque of the fibre. Use the torsional constant per degree to get the angle in degrees.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
In a radial field, the torque on the coil is \(NIAB\). It is balanced by the restoring torque \(k\theta\).

Step 2: Substitute
Take one turn: \(IAB=250\times10^{-6}\times0.04\times0.02=2\times10^{-7}\text{ Nm}\).

Step 3: Calculation
\(\theta=\dfrac{IAB}{k}=\dfrac{2\times10^{-7}}{5\times10^{-9}}=40\)
As \(k\) is per degree, \(\theta=40^{\circ}\).

Step 4: Compare the options
This is the second option.

Final Answer:
The coil rotates through 40 degrees. \[ \boxed{\text{(B) }40^{\circ}} \]