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 solve for the distances from the center of the wire's cross-section at which the magnetic field is half of its maximum value, we first consider the magnetic field behavior for different sections of the wire.

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

The magnetic field \(B\) at any point inside the wire is given by:

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

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

The maximum magnetic field inside the wire occurs at the 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 value is:

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

Setting the general expression for \( B \) equal to half the maximum:

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

Simplifying, we find:

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

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

The magnetic field \(B\) at any point outside the wire is given by:

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

The maximum magnetic field outside is on the surface \(r = a\):

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

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:

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

Simplifying, we find:

\( r = 2a \)

Therefore, the distances from the center where the magnetic field is half its maximum are \(\frac{a}{2}\) inside the wire and \(2a\) outside the wire.

Answer 2

Step 1: Understand the problem setup.
We have a long straight wire of circular cross-section with radius \( a \), carrying a steady current \( I \). The current is uniformly distributed across the cross-section. We need to find the distances from the center (inside and outside the wire) where the magnetic field is half of its maximum value.

Step 2: Expression for magnetic field inside the wire (r < a).
For a uniformly distributed current, the magnetic field inside the wire at a distance \( r \) from the center is given by:
\[ B_{\text{inside}} = \frac{\mu_0 I r}{2 \pi a^2}. \] The maximum magnetic field occurs at the surface of the wire (\( r = a \)) and is:
\[ B_{\text{max}} = \frac{\mu_0 I}{2 \pi a}. \] We are asked to find the point where \( B_{\text{inside}} = \frac{1}{2} B_{\text{max}} \):
\[ \frac{\mu_0 I r}{2 \pi a^2} = \frac{1}{2} \times \frac{\mu_0 I}{2 \pi a}. \] Simplify this equation:
\[ \frac{r}{a^2} = \frac{1}{2a} \quad \Rightarrow \quad r = \frac{a}{2}. \] Hence, inside the wire, the magnetic field is half of its maximum value at \( r = \frac{a}{2} \).

Step 3: Expression for magnetic field outside the wire (r > a).
Outside the wire, the magnetic field behaves as if the entire current \( I \) were concentrated at the center. The magnetic field at distance \( r \) is:
\[ B_{\text{outside}} = \frac{\mu_0 I}{2 \pi r}. \] We need \( B_{\text{outside}} = \frac{1}{2} B_{\text{max}} \):
\[ \frac{\mu_0 I}{2 \pi r} = \frac{1}{2} \times \frac{\mu_0 I}{2 \pi a}. \] Simplify this equation:
\[ \frac{1}{r} = \frac{1}{2a} \quad \Rightarrow \quad r = 2a. \] Hence, outside the wire, the magnetic field is half of its maximum value at \( r = 2a \).

Step 4: Final Answer.
The distances from the center where the magnetic field is half of its maximum value are:
\[ \boxed{r = \frac{a}{2} \text{ (inside)}, \quad r = 2a \text{ (outside)}}. \]

Final Answer:
\( \frac{a}{2}, 2a \)