Question

A steady current $I$ flow through a long straight wire of radius ‘$a$’. The current is uniformly distributed across its cross section. The ratio of the magnetic fields due to the wire at distance $\frac{a}{4}$ and $3a$ respectively from the axis of the wire is

• A. 3 : 4
• B. 4 : 3
• C. 2 : 3
• D. 1 : 4

Hint:

Remember the general behavior: inside a wire, $B \propto r$, and outside a wire, $B \propto \frac{1}{r}$. Sketching this classic peak profile helps you quickly double-check your fractional ratios during exams!

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
According to Ampere’s Circuital Law, the magnetic field produced by a thick, current-carrying wire varies depending on whether the observation point is inside or outside the wire's physical boundary. Inside the wire, the field grows linearly with distance from the central axis because it only encloses a fraction of the total current. Outside the wire, the field decreases inversely with distance, as if all the current were concentrated along the central axis.

Step 2: Key Formula or Approach:
For a solid wire of radius $a$ carrying a uniformly distributed total current $I$: 1. Inside the wire ($r \le a$): $B_{\text{in}} = \frac{\mu_0 I r}{2\pi a^2}$ 2. Outside the wire ($r \ge a$): $B_{\text{out}} = \frac{\mu_0 I}{2\pi r}$

Step 3: Detailed Explanation:
Let's evaluate the magnetic field at both specified distances from the central axis: 1. At the first point ($r_1 = \frac{a}{4}$), which lies inside the wire ($r_1 & Lt; a$): \[ B_1 = \frac{\mu_0 I \left(\frac{a}{4}\right)}{2\pi a^2} = \frac{\mu_0 I}{8\pi a} \] 2. At the second point ($r_2 = 3a$), which lies outside the wire ($r_2 > a$): \[ B_2 = \frac{\mu_0 I}{2\pi (3a)} = \frac{\mu_0 I}{6\pi a} \] Now, find the ratio of these two magnetic fields ($B_1 : B_2$): \[ \frac{B_1}{B_2} = \frac{\frac{\mu_0 I}{8\pi a}}{\frac{\mu_0 I}{6\pi a}} = \frac{6\pi a}{8\pi a} = \frac{6}{8} = \frac{3}{4} \] Thus, the required ratio of the magnetic fields is $3 : 4$.

Step 4: Final Answer:
The ratio of the magnetic fields at the given distances is 3 : 4.