Q43 (2026)
The magnetic field at the centre of a current carrying circular loop of radius R is $16 \,\mu\mathrm{T}$ . The magnetic field at a distance $x = \sqrt{3} R$ on its axis from the centre is ____ $\mu\mathrm{T}$ .
- $2\sqrt{2}$
- $4$
- $2$
- $8$
### Related Formula
$$B_{\text{centre}} = \frac{\mu_0 I}{2R}$$
$$B_{\text{axis}} = \frac{\mu_0 I R^2}{2(x^2 + R^2)^{3/2}}$$
### Core Logic
Substitute the expression for $B_{\text{centre}}$ into the equation for $B_{\text{axis}}$ by evaluating the denominator with $x = \sqrt{3}R$.
### Step 1: Using Center Field
$$B_{\text{centre}} = \frac{\mu_0 I}{2R} = 16 \,\mu\mathrm{T}$$
### Step 2: Calculating Axial Field
Substitute $x = \sqrt{3}R$ into the axial field formula:
$$B_{\text{axis}} = \frac{\mu_0 I R^2}{2( (\sqrt{3}R)^2 + R^2 )^{3/2}}$$
$$= \frac{\mu_0 I R^2}{2(3R^2 + R^2)^{3/2}}$$
$$= \frac{\mu_0 I R^2}{2(4R^2)^{3/2}}$$
### Step 3: Final Deduction
$$= \frac{\mu_0 I R^2}{2(8R^3)} = \frac{\mu_0 I}{16R}$$
$$= \frac{1}{8} \left( \frac{\mu_0 I}{2R} \right) = \frac{1}{8} \times 16 \,\mu\mathrm{T} = 2 \,\mu\mathrm{T}$$
### Pattern Recognition
A classic standard derivation: at $x = \sqrt{3}R$, the field is $1/8$ of the center field. At $x = R$, it's $1/(2\sqrt{2})$. Committing these to memory saves 60 seconds.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
