Login/Sign Up
Home » 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.
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.
CBSE | Class 12 | Moving Charges and Magnetism | NCERT | Physics
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.

Radius of the coil X, r1 is given as 0.16 m

Number of turns in coil X, n1 is given as 20

Current in the coil X, I1 is given as 16 A

Radius of the coil Y, r2 is given as 0.1 m

Number of turns in coil Y, n2 is given as 25

Current in the coil Y, I2 is given as 18 A

The expression for magnetic field generated by the coil X at the centre is:

{B_1} = \frac{{{\mu _o}{N_1}{I_1}}}{{2{r_1}}}

Here, μ0 is the permeability of the free space having value 4\pi \times {10^{ - 7}}Tm/A

{B_1} = \frac{{4\pi \times {{10}^{ - 7}} \times 20 \times 16}}{{2 \times 0.16}}

= 4\pi \times {10^{ - 4}}T(Towards east)

The magnetic field generated by the coil Y at the centre can be calculated as:

{B_2} = \frac{{{\mu _0}{N_2}{I_2}}}{{2{r_2}}}

{B_2} = \frac{{4\pi \times {{10}^{ - 7}} \times 25 \times 18}}{{2 \times 0.10}}

= 9\pi \times {10^{ - 4}}T(Towards west)

Calculating the net magnetic field,

B = {B_2} - {B_1}

B = 9\pi \times {10^{ - 4}}T - 4\pi \times {10^{ - 4}}T

B = 5\pi \times {10^{ - 4}}T

B = 5 \times 3.14 \times {10^{ - 4}}

B = 1.57 \times {10^{ - 3}}T(Towards west)

i

More Material