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: Understanding the Concept:
A circular coil of radius \(R\) carrying current \(I\) produces a magnetic field at its centre along its axis. The two coils are perpendicular, so their axes are perpendicular too.
Coil A lies in the XY plane, so its axis is the Z axis. Coil B lies in the YZ plane, so its axis is the X axis.

Step 2: Key Formula or Approach:
The field at the centre of a coil is:
\[ B = \frac{\mu_0 I}{2R} \]
Because \(\vec{B}_A\) and \(\vec{B}_B\) are perpendicular, the net field has magnitude \(B = \sqrt{B_A^2 + B_B^2}\).

Step 3: Field of coil A:
Here \(R = 10\ \text{cm} = 0.1\ \text{m}\) and \(I_A = 1\ \text{A}\).
\[ B_A = \frac{4\pi \times 10^{-7} \times 1}{2 \times 0.1} = 2\pi \times 10^{-6}\ \text{T} \approx 6.28 \times 10^{-6}\ \text{T} \]

Step 4: Field of coil B:
The current is \(I_B = \sqrt{3}\ \text{A}\), so the field is \(\sqrt{3}\) times larger.
\[ B_B = \sqrt{3} \times 2\pi \times 10^{-6} \approx 1.088 \times 10^{-5}\ \text{T} \]

Step 5: Net field:
\[ B = \sqrt{(2\pi \times 10^{-6})^2 + (\sqrt{3} \times 2\pi \times 10^{-6})^2} = 2\pi \times 10^{-6}\sqrt{1 + 3} = 4\pi \times 10^{-6}\ \text{T} \]
This is about \(1.26 \times 10^{-5}\ \text{T}\). The direction makes an angle \(\tan^{-1}\sqrt{3} = 60^{\circ}\) with the axis of coil A.

Final Answer:
The net magnetic field at the common centre has magnitude \(4\pi \times 10^{-6}\ \text{T}\). \[ \boxed{B \approx 1.26 \times 10^{-5}\ \text{T}} \]