Two similar circular loops of radius R are lying concentrically with their planes at right angles to each other. The currents flowign in them are `l ` and `lsqrt(3)` respectively. The magnetic field at the centre of the coil is

Options

A. `(mu_(0)l)/(2R)`
B. `(mu_(0)l)/R`
C. `(4mu_(0)l)/R`
D. `(sqrt(3)mu_(0)l)/R`

Answer

To find the magnetic field at the center of two similar circular loops of radius \( R \) carrying currents \( I \) and \( I\sqrt{3} \) respectively, we can follow these steps:

### Step 1: Determine the magnetic field due to each loop

The magnetic field \( B \) at the center of a circular loop carrying current \( I \) is given by the formula:
\[ B = \frac{\mu_0 I}{2R} \]
where \( \mu_0 \) is the permeability of free space and \( R \) is the radius of the loop.

### Step 2: Calculate the magnetic field \( B_1 \) for the first loop

For the first loop with current \( I \):
\[ B_1 = \frac{\mu_0 I}{2R} \]

### Step 3: Calculate the magnetic field \( B_2 \) for the second loop

For the second loop with current \( I\sqrt{3} \):
\[ B_2 = \frac{\mu_0 I\sqrt{3}}{2R} \]

### Step 4: Relate \( B_2 \) to \( B_1 \)

We can express \( B_2 \) in terms of \( B_1 \):
\[ B_2 = \sqrt{3} \cdot B_1 \]

### Step 5: Calculate the net magnetic field at the center

Since the two magnetic fields \( B_1 \) and \( B_2 \) are perpendicular to each other, the net magnetic field \( B_{net} \) can be calculated using the Pythagorean theorem:
\[ B_{net} = \sqrt{B_1^2 + B_2^2} \]
Substituting \( B_2 = \sqrt{3} B_1 \):
\[ B_{net} = \sqrt{B_1^2 + (\sqrt{3} B_1)^2} \]
\[ B_{net} = \sqrt{B_1^2 + 3B_1^2} = \sqrt{4B_1^2} = 2B_1 \]

### Step 6: Substitute \( B_1 \) into the equation

Now substituting \( B_1 = \frac{\mu_0 I}{2R} \):
\[ B_{net} = 2 \left(\frac{\mu_0 I}{2R}\right) = \frac{\mu_0 I}{R} \]

### Final Answer

Thus, the net magnetic field at the center of the loops is:
\[ B_{net} = \frac{\mu_0 I}{R} \]