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 :
- \( \frac{\sqrt{2} \mu_0 I}{a} \)
- \( \frac{\mu_0 I}{2a} \)
- \( \frac{\mu_0 I}{\sqrt{2} a} \)
- \( \frac{2 \mu_0 I}{a} \)
The Correct Option is C
Approach Solution - 1
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 \( \frac{\mu_0 I}{\sqrt{2}a} \).
Approach Solution -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:
\[ B_{\text{net}} = \sqrt{B^2 + B^2} = \frac{\mu_0 I}{\sqrt{2}a} \]
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