Question

A 50-turn circular coil of radius 2.0 cm carrying a current of 5.0 A is rotated in a magnetic field of strength 0.20 T. (a) What is the maximum torque that acts on the coil? (b) In a particular position of the coil, the torque acting on it is half of this maximum. What is the angle between the magnetic field and the plane of the coil?

Answer 1

Given:
No. of turns of the coil, n = 50
Magnetic field intensity, B = 0.20 T = 2 × 10⁻¹ T
Radius of the coil, r = 0.02 m = 2 × 10⁻² m
Magnitude of current =5 A
Torque acting on the coil,
τ = niABsinθ
Here, A is the area of the coil and θ is the angle between the area vector and the magnetic field.
τ is maximum when θ = 90°.
τ_max = niABsin90°
= 50 × 5 × 3.14 × 4 × 10⁻⁴ × 2 × 10⁻¹
= 6.28 × 10⁻² N-m
$$\begin{gathered} \mathrm{Given},\tau =\frac{1}{2}\times {\tau }_{max} \\ \Rightarrow \mathrm{sin\theta }=\frac{1}{2} \\ \Rightarrow \theta \mathit{}=30° \end{gathered}$$
So, the angle between the magnetic field and the plane of the coil = 90° − 30° = 60°