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

The problem asks for the magnitude of the net magnetic induction at the common center 'O' of two identical, perpendicular, current-carrying circular loops.

Concept Used:

The solution is based on two key principles of magnetism:

1. Magnetic Field at the Center of a Circular Loop: The magnitude of the magnetic field (magnetic induction) at the center of a circular loop of radius 'a' carrying a current 'I' is given by the Biot-Savart law as:

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

The direction of this magnetic field is perpendicular to the plane of the loop and is determined by the Right-Hand Thumb Rule.

2. Principle of Superposition: The net magnetic field at any point due to a system of multiple current-carrying conductors is the vector sum of the magnetic fields produced by each conductor individually at that point.

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

Step-by-Step Solution:

Step 1: Determine the magnetic field due to loop A.

Loop A is a horizontal circular loop. The current 'I' is in the anti-clockwise direction when viewed from above. According to the Right-Hand Thumb Rule (curling the fingers in the direction of the current), the thumb points upwards. Therefore, the magnetic field \( \vec{B}_A \) produced by loop A at the center O is directed vertically upwards. Let's consider the vertical direction to be along the z-axis.

The magnitude of this field is:

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

So, the vector representation is \( \vec{B}_A = \frac{\mu_0 I}{2a} \hat{k} \).

Step 2: Determine the magnetic field due to loop B.

Loop B is a vertical circular loop, perpendicular to loop A. The current 'I' is also anti-clockwise as shown. Let's assume loop B lies in the y-z plane. An anti-clockwise current in this plane (as seen from the positive x-axis) would produce a magnetic field along the positive x-axis, according to the Right-Hand Thumb Rule. Thus, the magnetic field \( \vec{B}_B \) produced by loop B at the center O is directed horizontally.

The magnitude of this field is the same as that for loop A since the radius and current are identical:

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

So, the vector representation is \( \vec{B}_B = \frac{\mu_0 I}{2a} \hat{i} \).

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

The net magnetic field is the vector sum of the individual fields:

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

Since \( \vec{B}_A \) is along the z-axis and \( \vec{B}_B \) is along the x-axis, the two magnetic field vectors are perpendicular to each other.

Final Computation & Result:

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

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

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

Substituting the 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 centre is \( \frac{\sqrt{2}\mu_0 I}{2a} \) or equivalently \( \frac{\mu_0 I}{\sqrt{2}a} \).

Answer 2

Calculate the magnetic field at the center due to one loop:

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

Since there are two loops in perpendicular planes, the resultant magnetic field is:

\[ B_{\text{net}} = \sqrt{B^2 + B^2} = \frac{\mu_0 I}{\sqrt{2}a} \]