Question

Three rings, each with equal radius \( r \), are placed mutually perpendicular to each other and each having centre at the origin of the coordinate system. If \( I \) is the current passing through each ring, the magnetic field value at the common centre is:

• A. \( 0 \)
• B. \( (\sqrt{3} - 1) \frac{\mu_0 I}{2\pi r} \)
• C. \( \frac{\sqrt{3} \mu_0 I}{2r} \)
• D. \( \frac{\sqrt{2} \mu_0 I}{2r} \)

Hint:

When multiple mutually perpendicular rings carry current, their magnetic fields add vectorially using: \[ B_{\text{net}} = \sqrt{B_x^2 + B_y^2 + B_z^2}. \] For three mutually perpendicular identical rings, use: \[ B_{\text{net}} = \frac{\sqrt{3} \mu_0 I}{2r}. \]

Answer 1

The Correct Option is C

To determine the magnetic field at the center of three mutually perpendicular rings, each with radius \( r \) and carrying a current \( I \), we use the Biot-Savart Law. The Biot-Savart Law provides the magnetic field contribution \( d\mathbf{B} \) from an element of current-carrying wire:

\( d\mathbf{B} = \frac{\mu_0}{4\pi} \frac{I d\mathbf{l} \times \mathbf{r}}{r^3} \)

For a complete circular ring, the magnetic field at the center due to symmetry is perpendicular to the plane of the ring, and its magnitude is given by:

\( \mathbf{B} = \frac{\mu_0 I}{2r} \)

Since there are three rings, each mutually perpendicular, the total magnetic field must consider vector addition. Each ring contributes a magnetic field of magnitude:

\( \mathbf{B_1} = \mathbf{B_2} = \mathbf{B_3} = \frac{\mu_0 I}{2r} \)

The net magnetic field \( \mathbf{B}_{net} \) is the vector sum:

\( \mathbf{B}_{net} = \sqrt{\mathbf{B_1}^2 + \mathbf{B_2}^2 + \mathbf{B_3}^2} \)

This results in:

\( \mathbf{B}_{net} = \sqrt{\left(\frac{\mu_0 I}{2r}\right)^2 + \left(\frac{\mu_0 I}{2r}\right)^2 + \left(\frac{\mu_0 I}{2r}\right)^2} \)

\( = \sqrt{3\left(\frac{\mu_0 I}{2r}\right)^2} \)

\( = \frac{\sqrt{3} \mu_0 I}{2r} \)

Therefore, the magnetic field at the common center is:

\( \frac{\sqrt{3} \mu_0 I}{2r} \)

Answer 2

Step 1: Magnetic Field at the Centre of a Single Current-Carrying Ring The magnetic field at the center of a circular current loop carrying current \( I \) is given by: \[ B = \frac{\mu_0 I}{2r}. \] Since each ring is mutually perpendicular to each other, the individual magnetic fields act along three coordinate axes \( (x, y, z) \). The net magnetic field at the center is the vector sum of the contributions from all three rings.
Step 2: Adding Magnetic Fields Vectorially Since the three rings are mutually perpendicular, the resultant magnetic field follows: \[ B_{\text{net}} = \sqrt{B_x^2 + B_y^2 + B_z^2}. \] Substituting \( B_x = B_y = B_z = \frac{\mu_0 I}{2r} \): \[ B_{\text{net}} = \sqrt{\left(\frac{\mu_0 I}{2r}\right)^2 + \left(\frac{\mu_0 I}{2r}\right)^2 + \left(\frac{\mu_0 I}{2r}\right)^2}. \] \[ B_{\text{net}} = \sqrt{3} \times \frac{\mu_0 I}{2r}. \] Thus, the correct answer is: \[ \boxed{\frac{\sqrt{3} \mu_0 I}{2r}}. \]