Question

Consider a long straight wire of a circular cross-section (radius \( a \)) carrying a steady current \( I \). The current is uniformly distributed across this cross-section. The distances from the center of the wire's cross-section at which the magnetic field (inside the wire, outside the wire) is half of the maximum possible magnetic field, anywhere due to the wire, will be:

• A. \( \frac{a}{2}, 3a \)
• B. \( \frac{a}{4}, 2a \)
• C. \( \frac{a}{2}, 2a \)
• D. \( \frac{a}{4}, \frac{3a}{2} \)

Hint:

The magnetic field inside a wire increases linearly with distance from the center, while outside the wire, it follows the same pattern as that for a point charge.

Answer 1

The Correct Option is C

To determine the radial positions where the magnetic field strength is half of its maximum, we analyze the wire's interior and exterior separately.

Within the Wire \((r \leq a)\):

The magnetic field \(B\) at a radial distance \(r\) is:

\( B = \frac{\mu_0 I r}{2\pi a^2} \)

where \( \mu_0 \) is the permeability of free space, \( I \) is the total current, \( r \) is the radial distance, and \( a \) is the wire's radius.

The maximum magnetic field inside the wire is located at its surface, \( r = a \):

\( B_{\text{max, inside}} = \frac{\mu_0 I a}{2\pi a^2} = \frac{\mu_0 I}{2\pi a} \)

Half of this maximum is:

\( \frac{B_{\text{max, inside}}}{2} = \frac{\mu_0 I}{4 \pi a} \)

Equating the general expression for \( B \) inside to half the maximum:

\( \frac{\mu_0 I r}{2\pi a^2} = \frac{\mu_0 I}{4 \pi a} \)

Solving for \( r \) yields:

\( r = \frac{a}{2} \)

Outside the Wire \((r > a)\):

The magnetic field \(B\) at a radial distance \(r\) is:

\( B = \frac{\mu_0 I}{2\pi r} \)

The maximum magnetic field outside occurs at \(r = a\):

\( B_{\text{max, outside}} = \frac{\mu_0 I}{2\pi a} \)

Half of this maximum is:

\( \frac{B_{\text{max, outside}}}{2} = \frac{\mu_0 I}{4 \pi a} \)

Equating the general expression for \( B \) outside to half the maximum:

\( \frac{\mu_0 I}{2\pi r} = \frac{\mu_0 I}{4\pi a} \)

Solving for \( r \) yields:

\( r = 2a \)

Consequently, the magnetic field is half its maximum value at distances of \(\frac{a}{2}\) from the center (inside the wire) and \(2a\) from the center (outside the wire).