1

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?

Open in App

Solution

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

$\mathrm{Given},\tau =\frac{1}{2}\times {\tau}_{max}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{sin\theta}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \theta \mathit{}=30\xb0$

So, the angle between the magnetic field and the plane of the coil = 90° − 30° = 60°

No. of turns of the coil, n = 50

Magnetic field intensity, B = 0.20 T = 2 × 10

Radius of the coil, r = 0.02 m = 2 × 10

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

= 50 × 5 × 3.14 × 4 × 10

= 6.28 × 10

$\mathrm{Given},\tau =\frac{1}{2}\times {\tau}_{max}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{sin\theta}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \theta \mathit{}=30\xb0$

So, the angle between the magnetic field and the plane of the coil = 90° − 30° = 60°

0

View More

Join BYJU'S Learning Program

Join BYJU'S Learning Program