Question

Two concentric circular coils $X$ and $Y$ of radii $16\text{cm}$ and $10\text{cm}$ respectively lie in same vertical plane containing the north-south direction. Coil $X$ has $20$ turns and carries a current of $16\text{A}$, coil $Y$ has $25$ turns and carries a current of $18\text{A}$. The sense of current in $X$ is anticlockwise and in coil $Y$ it is clockwise, for an observer looking at the coils, facing the west.
The magnitude and direction of the magnetic field at their common centre is:
(Neglect horizontal component of Earth's magnetic field)

• A. $9\pi \times {10}^{-4}\text{T}$ towards the west
• B. $9\pi \times {10}^{-4}\text{T}$ towards the east
• C. $5\pi \times {10}^{-4}\text{T}$ towards the west
• D. $4\pi \times {10}^{-4}\text{T}$ towards the east

Answer 1

The correct option is C $5\pi \times {10}^{-4}\text{T}$ towards the west

Using right-hand thumb rule the direction of magnetic field due to coil $X$ and $Y$ is shown in figure below as $B_x$ and $B_y$.

So, the magnetic field due to coil $X$ at the centre is given by

$B_x=\frac{{\mu }_0{n}_x{i}_1}{2r_x}$

$B_x=\frac{(4\pi \times {10}^{-7})\times 20\times 16}{2\times (16\times {10}^{-2})}$

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

Similarly, the magnetic field at common centre due to coil $Y$,

$B_y=\frac{{\mu }_0{n}_y{i}_2}{2r_y}$

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

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

So, net magnetic field at the common centre is

$B_{net}=B_y-B_x(\because B_y>B_x)$

$B_{net}=(9\pi \times {10}^{-4})-(4\pi \times {10}^{-4})$

$B_{net}=5\pi \times {10}^{-4}\text{T}$ (Towards west)

Hence, option $\left(b\right)$ is correct.

Why this question ? Tip: In such problems always observe the plane in which circular wire is placed. Then by using right-hand thumb rule, curl your right hand along the direction of current and thumb will point the direction of magnetic field at centre (perpendicular to plane of wire).