Question:

A long straight wire of radius \( a \) carries a steady current \( I \). The current is uniformly distributed across its cross-section. The ratio of the magnetic field at \( a/2 \) and \( 2a \) from the axis of the wire is:

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Magnetic field inside a current-carrying wire varies linearly, while outside it follows an inverse relation with distance.
Updated On: May 22, 2025
  • \( 1:4 \)
  • \( 4:1 \)
  • \( 1:1 \)
  • \( 3:4 \)
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The Correct Option is C

Approach Solution - 1

Step 1: {Using Ampere’s Circuital Law}
The magnetic field due to a straight wire is given by: \[ B = \frac{\mu_0 I}{2\pi r} \] Step 2: {Finding the Magnetic Field at \( a/2 \) and \( 2a \)}
Magnetic field at \( r = a/2 \): \[ B_{a/2} = \frac{\mu_0 I}{2\pi (a/2)} \] \[ B_{a/2} = \frac{\mu_0 I}{\pi a} \] Magnetic field at \( r = 2a \): \[ B_{2a} = \frac{\mu_0 I}{2\pi (2a)} \] \[ B_{2a} = \frac{\mu_0 I}{4\pi a} \] Step 3: {Calculating the Ratio}
\[ \frac{B_{a/2}}{B_{2a}} = \frac{\frac{\mu_0 I}{\pi a}}{\frac{\mu_0 I}{4\pi a}} = \frac{1}{1} = 1:1 \] Thus, the correct answer is \( 1:1 \).
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Approach Solution -2

Step 1: Understand the problem
We are given a long straight wire of radius \( a \) carrying a steady current \( I \), uniformly distributed across its cross-section.
We need to find the ratio of magnetic field at two points:
- Point 1: at a distance \( r = \frac{a}{2} \) (inside the wire)
- Point 2: at a distance \( r = 2a \) (outside the wire)

Step 2: Magnetic field inside the wire (r < a)
For \( r < a \), the magnetic field is given by Ampère’s Law:
\( B_{\text{inside}} = \frac{\mu_0 I r}{2\pi a^2} \)
Substituting \( r = \frac{a}{2} \):
\( B_{a/2} = \frac{\mu_0 I (a/2)}{2\pi a^2} = \frac{\mu_0 I}{4\pi a} \)

Step 3: Magnetic field outside the wire (r > a)
For \( r > a \), the wire behaves like a thin current-carrying conductor. The magnetic field is:
\( B_{\text{outside}} = \frac{\mu_0 I}{2\pi r} \)
Substituting \( r = 2a \):
\( B_{2a} = \frac{\mu_0 I}{2\pi (2a)} = \frac{\mu_0 I}{4\pi a} \)

Step 4: Take the ratio
\( \frac{B_{a/2}}{B_{2a}} = \frac{\mu_0 I / 4\pi a}{\mu_0 I / 4\pi a} = 1 \)
So, the ratio is:
\( B_{a/2} : B_{2a} = 1 : 1 \)

Final Answer: \( 1:1 \)
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