Question

A circular coil of radius \(10\,\text{cm}\) is carrying a current. From the centre of the coil, the distance of a point on its axis where \[ \frac{dB}{dx} \] becomes maximum is \[ \left( \frac{dB}{dx} \text{ is the rate of change of magnetic field with distance }x \text{ from the centre of the coil} \right) \]

• A. \(40\,\text{cm}\)
• B. \(20\,\text{cm}\)
• C. \(10\,\text{cm}\)
• D. \(5\,\text{cm}\)

Hint:

For a circular current-carrying coil, \[ \boxed{ B=\frac{\mu_0IR^2}{2(R^2+x^2)^{3/2}}. } \] The magnitude of the field gradient is maximum at \[ \boxed{x=\frac{R}{2}.} \]

Answer 1

The Correct Option is D

Step 1: Write the magnetic field on the axis of a circular coil. The magnetic field at a distance \(x\) from the centre is \[ B=\frac{\mu_0IR^2}{2(R^2+x^2)^{3/2}}, \] where \(R\) is the radius of the coil.

Step 2: Differentiate with respect to \(x\). Differentiating, \[ \frac{dB}{dx} = -\frac{3\mu_0IR^2x} {2(R^2+x^2)^{5/2}}. \] Ignoring the constant factors, we maximize \[ f(x)=\frac{x}{(R^2+x^2)^{5/2}}. \]

Step 3: Find the point where \(\dfrac{dB}{dx}\) is maximum. Differentiating, \[ \frac{df}{dx} = \frac{R^2-4x^2} {(R^2+x^2)^{7/2}}. \] Setting \[ \frac{df}{dx}=0, \] gives \[ R^2-4x^2=0, \] or \[ x=\frac{R}{2}. \] Since \[ R=10\,\text{cm}, \] \[ x=\frac{10}{2}=5\,\text{cm}. \] Hence, \[ \boxed{x=5\,\text{cm}.} \] Therefore, the correct option is \(\boxed{(D)}\).