A circular coil of 200 turns, radius 5 cm carries a current of 2.5 A. It is suspended vertically in a uniform horizontal magnetic field of 0.25 T, with the plane of the coil making an angle of 60° with the field lines. Calculate the magnitude of the torque that must be applied on it to prevent it from turning.

Solution

Given,
Number of turns in the coil, N = 200
Radius of coil, r = 5 $\times 10^{- 2} m$
Area of coil, $A = {πr}^{2}$
$= \frac{22}{7} \times 5 \times 10^{- 2} \times 5 \times 10^{- 2} = 7.587 \times 10^{- 3} m^{2}$
Current carried by the coil, I = 2.5 A
Magnetic field, B = 0.25 T
Angle made by coil with the field lines = 60^o

Using the formula for torque,
$τ = NBIA \cos θ$
= 200 x 0.25 x 2.5 x 7.857 x 10^–3 x cos 60° Nm
= 0.49 Nm

An opposite and equal torque is required in order to prevent the coil from turning. Thus, the magnitude of the applied torque should be 0.49 m.

Some More Questions From Moving Charges And Magnetism Chapter

Two moving coil meters, M₁ and M₂ have the following particulars:
R₁ = 10 Ω N₁ = 30,
A₁ = 3.6 x 10^–3 m², B₁ = 0.25 T
R₂ = 14 Ω; N₂ = 42,
A₂ = 1.8 x 10^–3 m², B₂ = 0.50 T
(The spring constants k are identical for the two meters).
Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M₂ and M_1.

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Moving Charges And Magnetism

Some More Questions From Moving Charges And Magnetism Chapter

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