Two concentric coils X and Y of radii 16 cm and 10 cm respectively lie in the same vertical plane containing the north to south direction. Coil X has 20 turns and carries a current of 16 A ; coil Y has 25 turns and carries a current of 18 A. The sense of current in X is anti clockwise and in Y, clockwise, for an observer looking at the coil facing west. Give the magnitude and direction of the net magnetic field due to the coils at their centre.

Solution

Consider X axis in East-West directions and YY’ in North-South direction respectively.
Plane of coil is in Y-Z axis plane.
Direction of magnetic field due to coil X is in East, and due to coil Y is in west direction.

Given, for coil X
Radius of coil, r_x = 16 cm = 0.16 m
Number of turns, N_x = 20
Current passing through coil, I_x = 16 A

Magnetic field at the centre of coil X is

B_{x} = \frac{μ_{0} I_{x} N_{x}}{2 r_{x}} = \frac{4 π \times 10^{- 7} \times 16 \times 20}{2 \times 0.16} = 4 π \times 10^{- 4} T

The current in the coil X is anticlockwise therefore, the field B_x is directed towards east.

Given, for coil Y
Radius, r_y = 10 cm = 0.10 m
Number of turns, N_y = 25
Current pasing through, I_y = 18 A

Magnetic field at the centre of coil Y is

$B_{y} = \frac{μ_{0} I_{y} N_{y}}{2 r_{y}} = \frac{4 π \times 10^{- 7} \times 18 \times 25}{2 \times 0.10} = 9 π \times 10^{- 4} T$

The direction of magnetic field induction B_y is towards west.

Hence,

Net magnetic field = B_y – B_x= 9 $π$ x 10^–4 – 4 $π$ x 10^–4= 5 $π$ x 10^–4= 1.6 x 10^–3 T (Towards west).

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

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