**Question:**

Two concentric circular 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 the current in X is anticlockwise, and clockwise in Y, for an observer looking at the coils facing west. Give the magnitude and direction of the net magnetic field due to the coils at their centre.

**Solution:**

Radius of coil X, *r*1 = 16 cm = 0.16 m

Radius of coil Y,* r*2 = 10 cm = 0.1 m

Number of turns of on coil X, *n*1 = 20

Number of turns of on coil Y, *n*2 = 25

Current in coil X, *I*1 = 16 A

Current in coil Y, *I*2 = 18 A

Magnetic field due to coil X at their centre is given by the relation,

$B_{1}=\frac{\mu_{0} n_{1} I_{1}}{2 r_{1}}$

Where,

$\mu_{0}=$ Permeability of free space $=4 \pi \times 10^{-7} \mathrm{~T} \mathrm{~m} \mathrm{~A}^{-1}$

$\therefore B_{1}=\frac{4 \pi \times 10^{-7} \times 20 \times 16}{2 \times 0.16}$

$=4 \pi \times 10^{-4} \mathrm{~T}$ (towards East)

Magnetic field due to coil Y at their centre is given by the relation,

B_{2}=\frac{\mu_{0} n_{2} I_{2}}{2 r_{2}}

$=\frac{4 \pi \times 10^{-7} \times 25 \times 18}{2 \times 0.10}$

$=9 \pi \times 10^{-4} \mathrm{~T}$ (towards West)

Hence, net magnetic field can be obtained as:

$B=B_{2}-B_{1}$

$=9 \pi \times 10^{-4}-4 \pi \times 10^{-4}$

$=5 \pi \times 10^{-4} \mathrm{~T}$

$=1.57 \times 10^{-3} \mathrm{~T}$ (towards West)

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