A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30° with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?

Solution

Here,
Length of the side of the square, l = 10 cm = 0.10 m
Number of turns in the coil, N = 20
Current carried by the conductor, I = 12 A
Angle made by the coil with the magnetic field, $θ = 30 °$
Magnetic field , B = 0.80 T
Area of the conductor $A = l \times l = 0.1 \times 0.1 = 0.01 m^{2}$
Therefore,

Torque is given by, $τ = NBIA \sin θ = 20 \times 0.80 \times 12 \times (0.1)^{2} \times \sin 90 ° = 0.96 Nm$

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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