Question

The dipole moment of a circular loop carrying a current $I$, is $m$ and the magnetic field at the centre of the loop is $B_1$. When the dipole moment is doubled by keeping the current constant, the magnetic field at the centre of the loop is $B_2$. The ratio $\frac{B_1}{B_2}$ is :

• A. $2$
• B. $\sqrt{3}$
• C. $\sqrt{2}$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

The Correct Option is C

The question involves understanding magnetic dipole moments and the magnetic field in a circular loop. Let's analyze each step to find the solution.

1. **Magnetic Dipole Moment (m):**
The magnetic dipole moment for a circular loop carrying a current \( I \) and having a radius \( r \) is given by:

m = I \cdot A, where A is the area of the loop. For a circular loop, A = \pi r^2.

2. **Magnetic Field at the Centre (B):**
The magnetic field at the center of the loop is given by:

B = \frac{\mu_0 I}{2r}, where \mu_0 is the permeability of free space.

3. **Relation Between m and B at the Centre:** B = \frac{\mu_0 \cdot m}{2 \pi r^3}. However, since m = I \cdot \pi r^2, we can rearrange to find:

B = \frac{\mu_0 \cdot I}{2r} = \frac{\mu_0 \cdot m}{2 \pi r^3}

4. **Doubling the Dipole Moment:** If the dipole moment \( m \) is doubled while keeping \( I \) constant, it means the area or size of the loop is changed. Let the new dipole moment be m_2 = 2m. We now analyze the change in magnetic field.

5. **New Magnetic Field (B2):** Using the same relation from step 3, the new magnetic field B_2 is proportional to the new dipole moment:

B_2 = \frac{\mu_0 \cdot m_2}{2 \pi r_2^3}

Given m_2 = 2m and realizing that the loop size is such that its new radius r_2 = \sqrt{2} \cdot r, we can substitute and rearrange:

B_2 = \frac{\mu_0 \cdot (2m)}{2 \pi (\sqrt{2}r)^3} = \frac{\mu_0 \cdot m}{\pi 2.828r^3}

6. **Finding the Ratio:** To find the ratio \frac{B_1}{B_2}:

\frac{B_1}{B_2} = \frac{\frac{\mu_0 \cdot m}{2 \pi r^3}}{\frac{\mu_0 \cdot m}{\pi 2.828r^3}}

Simplifying, we find:

\frac{B_1}{B_2} = \sqrt{2}

Hence, the correct answer is \sqrt{2}.