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 $16A$ ; coil $Y$ has $25$ turns and carries a current of $18A$ . 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.

Answer 1

To answer this question, we must know about Maxwell’s Right hand thumb rule and know how the magnetic field at the center of a current carrying coil is given.First we will calculate the magnetic field at the center of x and then at the center of y and then by using Maxwell’s Right hand thumb rule we will find the net magnetic field due to the coils at their centre.

Formula used:
${B_x} = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{2{I_x}\pi {n_x}}}{{{R_x}}}$
Where, ${B_x}$ is the Magnetic field at the center of X, ${I_x}$ is the current in the coil X, ${n_x}$ is the number of turns and ${R_x}$ is the radius of the coil.

Complete step by step answer:
According to the question,
For coil $X$: Radius ${R_1} = 0.16\,m$, ${n_x} = 20$ and ${I_x} = 16\,A$.
Now we will calculate the Magnetic field at the center of X by putting the values in the formula,
$\because {B_x} = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{2{I_x}\pi {n_x}}}{{{R_x}}} \\\ \Rightarrow {B_x} = \dfrac{{{{10}^{ - 7}} \times 2 \times 16 \times 3.14 \times 20}}{{0.16}} \\\ \Rightarrow {B_x} = 12.56 \times {10^{ - 4}}T \\\$
Similarly, we will calculate the Magnetic field at the center of Y by putting the values.
For coil $Y$: Radius ${R_2} = 0.1\,m$, ${n_y} = 25$ and ${I_y} = 18\,A$.
$\because {B_y} = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{2{I_y}\pi {n_y}}}{{{R_t}}} \\\ \Rightarrow {B_y} = \dfrac{{{{10}^{ - 7}} \times 3.14 \times 18 \times 25}}{{0.1}} \\\ \Rightarrow {B_y} = 28.26 \times {10^{ - 4}}T \\\$
Now, we will calculate the net magnetic field. Here we can understand by Maxwell's right hand rule, ${B_x}$​ is towards the right (East direction) and from Maxwell's right hand rule, ${B_y}$ is towards the left (West direction). Hence, net magnetic field at the center is,
${B_y} - {B_x} = (28.26 \times {10^{ - 4}}) - (12.56 \times {10^{ - 4}}) \\\ \therefore {B_y} - {B_x} = 15.7 \times {10^{ - 4}}T \\\$
Hence, the net magnetic field at the centre is $15.7 \times {10^{ - 4}}T$.

Note: We must Remember the formula and Maxwell’s right hand thumb rule is used to find the direction of the magnetic field in the conductor. The magnetic field of lines originates from the north pole of the magnet and terminates on the south pole of the magnets. The magnetic poles always repel each other. Opposite magnetic poles always attract each other.