Question

A steady current 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:

Instead of dragging the constant variables through every step, memorize the proportionalities directly: Inside the wire, $B \propto r$, so at $r = \frac{a}{4}$, the relative intensity is $\frac{1}{4}$. Outside the wire, $B \propto \frac{1}{r}$, so at $r = 3a$, the relative intensity is $\frac{1}{3}$. Dividing the two relative fractions gives $\frac{1/4}{1/3} = \frac{3}{4}$ immediately!

Answer 1

The Correct Option is A

Concept: According to Ampere's Circuital Law, the magnetic field $B$ at a distance $r$ from the central axis of a long solid cylindrical wire of radius $a$ carrying a total current $I$ uniformly distributed across its cross-section depends on whether the point lies inside or outside the conductor:
• Inside the wire ($r & Lt; a$): The magnetic field increases linearly with distance from the central axis: \[ B_{\text{in}} = \frac{\mu_0 I r}{2\pi a^2} \]
• Outside the wire ($r \ge a$): The magnetic field behaves as if all the current were concentrated at the central axis, decreasing inversely with distance: \[ B_{\text{out}} = \frac{\mu_0 I}{2\pi r} \]

Step 1: Calculate the magnetic field at the interior point ($r_1 = \frac{a}{4}$).
Since $r_1 = \frac{a}{4}$ is less than the wire's radius $a$, this point is located inside the conductor's cross-section. Using the interior magnetic field formula: \[ B_1 = \frac{\mu_0 I \left(\frac{a}{4}\right)}{2\pi a^2} = \frac{\mu_0 I a}{8\pi a^2} = \frac{\mu_0 I}{8\pi a} \quad \cdots (1) \]

Step 2: Calculate the magnetic field at the exterior point ($r_2 = 3a$).
Since $r_2 = 3a$ is greater than the wire's radius $a$, this point is located outside the conductor. Using the exterior magnetic field formula: \[ B_2 = \frac{\mu_0 I}{2\pi (3a)} = \frac{\mu_0 I}{6\pi a} \quad \cdots (2) \]

Step 3: Determine the ratio of the two magnetic fields ($\frac{B_1}{B_2}$).
Dividing equation (1) by equation (2) to find the required ratio: \[ \frac{B_1}{B_2} = \frac{\frac{\mu_0 I}{8\pi a}}{\frac{\mu_0 I}{6\pi a}} \] Canceling out the common factors ($\mu_0, I, \pi, a$) from the numerator and denominator: \[ \frac{B_1}{B_2} = \frac{\frac{1}{8}}{\frac{1}{6}} = \frac{6}{8} \] Simplifying the fraction by dividing both numbers by their greatest common divisor ($2$): \[ \frac{B_1}{B_2} = \frac{3}{4} \] Therefore, the ratio of the magnetic fields is exactly $3 : 4$.