Question

Two insulated circular loop A and B radius ‘a’ carrying a current of ‘I’ in the anti clockwise direction as shown in figure. The magnitude of the magnetic induction at the centre will be :

• A. \( \frac{\sqrt{2} \mu_0 I}{a} \)
• B. \( \frac{\mu_0 I}{2a} \)
• C. \( \frac{\mu_0 I}{\sqrt{2} a} \)
• D. \( \frac{2 \mu_0 I}{a} \)

Answer 1

The Correct Option is C

This problem requires calculating the magnitude of the resultant magnetic field at the shared center 'O' of two identical, orthogonal, current-carrying circular loops.

Underlying Principles:

The solution relies on two fundamental concepts in magnetism:

1. Magnetic Field from a Circular Loop: The Biot-Savart law dictates that the magnetic field magnitude at the center of a circular loop with radius 'a' carrying current 'I' is:

\[ B = \frac{\mu_0 I}{2a} \]

The orientation of this field is perpendicular to the loop's plane, as determined by the Right-Hand Thumb Rule.

2. Superposition Principle: The total magnetic field at a point due to multiple current sources is the vector sum of the individual magnetic fields produced by each source at that point.

\[ \vec{B}_{net} = \vec{B}_1 + \vec{B}_2 + \dots \]

Solution Steps:

Step 1: Analyze the magnetic field generated by loop A.

Loop A is a horizontal loop with current 'I' flowing anti-clockwise when viewed from above. Applying the Right-Hand Thumb Rule, the thumb points upwards, indicating the magnetic field \( \vec{B}_A \) at center O is directed vertically. We define this as the z-axis.

The magnitude of this field is:

\[ B_A = \frac{\mu_0 I}{2a} \]

In vector form, this is \( \vec{B}_A = \frac{\mu_0 I}{2a} \hat{k} \).

Step 2: Analyze the magnetic field generated by loop B.

Loop B is a vertical loop, orthogonal to loop A. Its current 'I' also flows anti-clockwise. If we position loop B in the y-z plane, an anti-clockwise current viewed from the positive x-axis generates a magnetic field along the positive x-axis, per the Right-Hand Thumb Rule. Therefore, the magnetic field \( \vec{B}_B \) at center O is horizontal.

The magnitude is identical to that of loop A due to the same radius and current:

\[ B_B = \frac{\mu_0 I}{2a} \]

In vector form, this is \( \vec{B}_B = \frac{\mu_0 I}{2a} \hat{i} \).

Step 3: Compute the net magnetic field at center O.

The net magnetic field is the vector sum of \( \vec{B}_A \) and \( \vec{B}_B \):

\[ \vec{B}_{net} = \vec{B}_A + \vec{B}_B \]

Since \( \vec{B}_A \) is aligned with the z-axis and \( \vec{B}_B \) with the x-axis, these vectors are perpendicular.

Final Calculation & Result:

Step 4: Determine the magnitude of the net magnetic field.

The magnitude of the resultant of two perpendicular vectors is calculated using the Pythagorean theorem:

\[ |\vec{B}_{net}| = \sqrt{B_A^2 + B_B^2} \]

Substituting the field magnitudes:

\[ |\vec{B}_{net}| = \sqrt{\left(\frac{\mu_0 I}{2a}\right)^2 + \left(\frac{\mu_0 I}{2a}\right)^2} \] \[ |\vec{B}_{net}| = \sqrt{2 \left(\frac{\mu_0 I}{2a}\right)^2} \] \[ |\vec{B}_{net}| = \sqrt{2} \times \frac{\mu_0 I}{2a} \]

The magnitude of the magnetic induction at the center is \( \frac{\sqrt{2}\mu_0 I}{2a} \) or, alternatively, \( \frac{\mu_0 I}{\sqrt{2}a} \).