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?

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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°

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