Two identical circular coils, P and Q, carrying currents `1A` and `sqrt3A` respectively, are placed concentrically and perpendicular to each other lying in the XY and YZ plane. Find the magnitude and direction of the net magnetic field at the centre of the coils.

Explanation

To solve the problem of finding the magnitude and direction of the net magnetic field at the center of two identical circular coils P and Q carrying currents \(1A\) and \(\sqrt{3}A\) respectively, which are placed concentrically and perpendicular to each other, we can follow these steps:

### Step 1: Calculate the Magnetic Field due to Coil P
The magnetic field \(B_1\) at the center of a circular coil is given by the formula:
\[ B_1 = \frac{\mu_0 I_1}{2R} \]
where:
- \(I_1 = 1A\) (current in coil P)
- \(\mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A\) (permeability of free space)
- \(R\) is the radius of the coil (not given, but will cancel out later).

Substituting the values:
\[ B_1 = \frac{4\pi \times 10^{-7} \times 1}{2R} = \frac{2\pi \times 10^{-7}}{R} \, T \]

### Step 2: Calculate the Magnetic Field due to Coil Q
Similarly, for coil Q, the magnetic field \(B_2\) is given by:
\[ B_2 = \frac{\mu_0 I_2}{2R} \]
where:
- \(I_2 = \sqrt{3}A\) (current in coil Q).

Substituting the values:
\[ B_2 = \frac{4\pi \times 10^{-7} \times \sqrt{3}}{2R} = \frac{2\pi \sqrt{3} \times 10^{-7}}{R} \, T \]

### Step 3: Determine the Direction of the Magnetic Fields
- The magnetic field \(B_1\) due to coil P (in the XY plane) will be directed along the Z-axis (using the right-hand rule).
- The magnetic field \(B_2\) due to coil Q (in the YZ plane) will be directed along the X-axis.

### Step 4: Calculate the Magnitude of the Net Magnetic Field
Since \(B_1\) and \(B_2\) are perpendicular to each other, we can use the Pythagorean theorem to find the resultant magnetic field \(B_{net}\):
\[ B_{net} = \sqrt{B_1^2 + B_2^2} \]

Substituting the expressions for \(B_1\) and \(B_2\):
\[ B_{net} = \sqrt{\left(\frac{2\pi \times 10^{-7}}{R}\right)^2 + \left(\frac{2\pi \sqrt{3} \times 10^{-7}}{R}\right)^2} \]
\[ = \sqrt{\frac{(2\pi \times 10^{-7})^2}{R^2} + \frac{(2\pi \sqrt{3} \times 10^{-7})^2}{R^2}} \]
\[ = \frac{2\pi \times 10^{-7}}{R} \sqrt{1 + 3} = \frac{2\pi \times 10^{-7}}{R} \sqrt{4} = \frac{4\pi \times 10^{-7}}{R} \, T \]

### Step 5: Determine the Direction of the Resultant Magnetic Field
The angle \(\theta\) that the resultant magnetic field makes with \(B_1\) can be found using:
\[ \tan \theta = \frac{B_2}{B_1} = \frac{\frac{2\pi \sqrt{3} \times 10^{-7}}{R}}{\frac{2\pi \times 10^{-7}}{R}} = \sqrt{3} \]
Thus,
\[ \theta = \tan^{-1}(\sqrt{3}) = 60^\circ \]

### Final Answer
The magnitude of the net magnetic field at the center of the coils is:
\[ B_{net} = \frac{4\pi \times 10^{-7}}{R} \, T \]
And the direction of the net magnetic field is \(60^\circ\) with respect to the magnetic field due to coil P.