Question

A current \( I \) is uniformly distributed across a long straight non-magnetic wire (\( \mu_r = 1 \)) of circular cross-section with radius \( a \). Two points \( P \) and \( Q \) are at distances \( \frac{a}{3} \) and \( 9a \), respectively, from the axis of the wire. The ratio of the magnetic fields at points \( P \) and \( Q \) is ................

Hint:

The magnetic field around a long straight current-carrying wire is inversely proportional to the distance from the wire.

Answer 1

Correct Answer: 3

Step 1: Use Ampère’s law for magnetic field.
For a long straight wire with a uniformly distributed current, the magnetic field at a distance \( r \) from the wire is given by \[ B(r) = \frac{\mu_0 I}{2 \pi r}, \] where \( \mu_0 \) is the permeability of free space, \( I \) is the current, and \( r \) is the distance from the center of the wire. Step 2: Calculate the ratio of the magnetic fields.
The magnetic field at point \( P \) (at distance \( \frac{a}{3} \)) is \[ B_P = \frac{\mu_0 I}{2 \pi \frac{a}{3}} = \frac{3 \mu_0 I}{2 \pi a}. \] The magnetic field at point \( Q \) (at distance \( 9a \)) is \[ B_Q = \frac{\mu_0 I}{2 \pi 9a} = \frac{\mu_0 I}{18 \pi a}. \] Step 3: Calculate the ratio.
The ratio of the magnetic fields is \[ \frac{B_P}{B_Q} = \frac{\frac{3 \mu_0 I}{2 \pi a}}{\frac{\mu_0 I}{18 \pi a}} = \frac{3}{\frac{1}{18}} = 54. \] Final Answer: The ratio of the magnetic fields at points \( P \) and \( Q \) is \( \boxed{54}. \)