Question

The magnetic field due to a current carrying circular coil on its axis at a distance of \(\sqrt{2} d\) from the centre of the coil is B. If d is the diameter of the coil, then the magnetic field at the centre of the coil is

• A. 18B
• B. 27B
• C. 3B
• D. 9B

Hint:

When comparing magnetic fields at two different points for the same coil, it's best to write out the formulas for both and then take their ratio, or solve for the constant term (\(\mu_0 I / r\)) as done here. Be careful with algebra, especially with exponents like \( (r^2)^{3/2} = r^3 \).

Answer 1

The Correct Option is B

Step 1: Formula for magnetic field on the axis. The magnetic field at a distance \(x\) from the center on the axis of a circular coil of radius \(R\) is: \[ B_{axis} = \frac{\mu_0 N I R^2}{2(R^2 + x^2)^{3/2}} \] Given: \(x = \sqrt{2}d\). Since diameter \(d = 2R\), we have \(x = \sqrt{2}(2R) = 2\sqrt{2}R\).
Step 2: Substitute \(x\) into the formula. \[ x^2 = (2\sqrt{2}R)^2 = 8R^2 \] \[ R^2 + x^2 = R^2 + 8R^2 = 9R^2 \] Substitute this back into the field equation: \[ B_{axis} = \frac{\mu_0 N I R^2}{2(9R^2)^{3/2}} = \frac{\mu_0 N I R^2}{2(27R^3)} = \frac{\mu_0 N I}{54R} \] We are given \(B_{axis} = B\). So, \( B = \frac{\mu_0 N I}{54R} \).
Step 3: Formula for magnetic field at the center. At the center (\(x=0\)): \[ B_{center} = \frac{\mu_0 N I}{2R} \]
Step 4: Find the relation between \(B_{center}\) and \(B\). From Step 2, \(\frac{\mu_0 N I}{2R} = 27 \times \left( \frac{\mu_0 N I}{54R} \right)\). Therefore: \[ B_{center} = 27 \times B_{axis} = 27B \]