Question

A long straight wire of circular cross-section of radius \( a \) is carrying a steady current. The current is distributed uniformly across the cross-section of the wire. The ratio of the magnetic fields at points \( 0.5a \) and \( 1.5a \) from the centre of the wire is:

• A. \( 1:1 \)
• B. \( 2:3 \)
• C. \( 1:2 \)
• D. \( 3:4 \)

Hint:

For magnetic fields in current-carrying wires: - Inside the wire: \( B \propto r \). - Outside the wire: \( B \propto \frac{1}{r} \).

Answer 1

The Correct Option is B

Step 1: Magnetic Field Inside and Outside the Wire The magnetic field inside a current-carrying wire (for \( r<a \)) follows: \[ B_{\text{inside}} = \frac{\mu_0 I r}{2\pi a^2} \] For outside the wire (for \( r>a \)): \[ B_{\text{outside}} = \frac{\mu_0 I}{2\pi r} \] Step 2: Calculating the Ratio For \( r = 0.5a \): \[ B_{0.5a} = \frac{\mu_0 I (0.5a)}{2\pi a^2} = \frac{\mu_0 I}{4\pi a} \] For \( r = 1.5a \): \[ B_{1.5a} = \frac{\mu_0 I}{2\pi (1.5a)} = \frac{\mu_0 I}{3\pi a} \] Ratio: \[ \frac{B_{0.5a}}{B_{1.5a}} = \frac{\frac{\mu_0 I}{4\pi a}}{\frac{\mu_0 I}{3\pi a}} \] \[ = \frac{1}{4} \div \frac{1}{3} = \frac{3}{4} \] Thus, the correct ratio is: \[ 2:3 \]