Question

Two identical circular coils A and B of radius $10$ cm each are placed concentrically and mutually perpendicular in XY and YZ planes respectively. Currents of $1$ A and $\sqrt{3}$ A flow anticlockwise in the coils as seen from YZ plane. Find the magnitude of the net magnetic field at the common centre of the coils.

Hint:

Use \(B = \mu_0 I/2R\) for each coil. The two fields are perpendicular, so add them as \(\sqrt{B_A^2 + B_B^2}\).

Answer 1

Step 1: Set up the geometry:
Take the common centre as the origin. The normal to the XY plane is Z, and the normal to the YZ plane is X. The two fields at the origin therefore point along Z and along X, at right angles.

Step 2: Write the field in a common form:
For a coil the field at the centre is $B = \frac{\mu_0 I}{2R}$. Both coils have the same radius, so the common factor is:
\[ \frac{\mu_0}{2R} = \frac{4\pi \times 10^{-7}}{0.2} = 2\pi \times 10^{-6}\ \text{T A}^{-1} \]
Then each field is this factor times its current.

Step 3: Add the fields as vectors:
Two perpendicular vectors add by Pythagoras, and the common factor comes out of the root:
\[ B = \frac{\mu_0}{2R}\sqrt{I_A^2 + I_B^2} \]
Putting in the currents, $\sqrt{1^2 + (\sqrt{3})^2} = \sqrt{4} = 2$.
\[ B = 2\pi \times 10^{-6} \times 2 = 4\pi \times 10^{-6}\ \text{T} \]

Step 4: Check the number and the direction:
With $\pi = 3.14$, $B = 12.56 \times 10^{-6}\ \text{T}$, which is $1.256 \times 10^{-5}\ \text{T}$.
The ratio of the two fields is $B_B / B_A = \sqrt{3}$, so the resultant is tilted $60^{\circ}$ from the axis of coil A towards the axis of coil B.

Final Answer:
The net field is about $1.26 \times 10^{-5}$ T. \[ \boxed{B = 4\pi \times 10^{-6}\ \text{T} \approx 1.26 \times 10^{-5}\ \text{T}} \]