Question

Current \( I \) flows through a conducting wire of radius \( a \). The magnetic field \( B \) at a distance \( r \) from the centre of the wire (where \( r>a \) and \( \mu \) is the permeability of free space) is:

• A. \( \frac{\mu I}{2\pi a^2} \)
• B. \( \frac{\mu I r}{2\pi a^2} \)
• C. \( \frac{\mu I}{2\pi r} \)
• D. \( \frac{\mu I}{\pi r^2} \)

Hint:

The magnetic field around a long straight conductor is inversely proportional to the distance from the wire.

Answer 1

The Correct Option is C

To determine the magnetic field \(B\) at a distance \(r\) from the center of a conducting wire with current \(I\) flowing through it, we use Ampère's Law. According to Ampère's Law, the line integral of the magnetic field \(B\) around a closed path is equal to the permeability of free space \(\mu\) times the current enclosed by the path:

\[\oint B \cdot dl = \mu I_{enc}\] 

For a long straight wire, we consider a circular path of radius \(r\) centered on the wire. The magnetic field \(B\) is tangent to this circle and has uniform magnitude along the path, so:

\[B \cdot 2\pi r = \mu I\]

Solving for \(B\), we find:

\[B = \frac{\mu I}{2\pi r}\]

This yields the magnetic field at a distance \(r\) from the center of the wire, where \(r > a\). Hence, the correct expression for the magnetic field is:

\[\frac{\mu I}{2\pi r}\]

Answer 2

For a long straight conducting wire, the magnetic field at a distance \( r \) from the centre of the wire (for \( r>a \)) is given by Ampere’s law: \[ B = \frac{\mu I}{2\pi r} \] Where:
- \( B \) is the magnetic field at distance \( r \),
- \( I \) is the current in the wire,
- \( r \) is the distance from the wire,
- \( \mu \) is the permeability of free space.

Answer 3

The magnetic field outside a long straight current-carrying wire can also be found directly from the Biot-Savart law by integrating over the length of the wire, without invoking Ampere's law as a shortcut. For an infinitely long straight wire, this integration gives \( B = \frac{\mu I}{2\pi r} \) for any point at a perpendicular distance \( r \) from the wire's axis, valid once \( r \geq a \), since all the current is already enclosed there. Let's check each option against this result.

1. \( \frac{\mu I}{2\pi a^2} \): This expression has no dependence on \( r \) at all, which cannot be correct outside the wire since the field must weaken as we move farther away from the current.
2. \( \frac{\mu I r}{2\pi a^2} \): This grows linearly with \( r \), which is the form of the field found strictly inside a uniformly distributed current (\( r<a \)), where only a fraction of the current is enclosed. It does not apply once \( r>a \), since here the full current \( I \) is already enclosed and adding more distance cannot make the field grow.
3. \( \frac{\mu I}{2\pi r} \): This matches the Biot-Savart integration result exactly, and correctly shows the field weakening as \( 1/r \) once all of the current is enclosed.
4. \( \frac{\mu I}{\pi r^2} \): This falls off as \( 1/r^2 \), which is the pattern seen for fields from point sources or electric fields from a point charge, not the field circling a long straight wire, which always falls as \( 1/r \).

Only the option that decreases as \( 1/r \), with no leftover dependence on \( a \), matches the physics of a long straight wire once we are outside it.

Therefore, the correct answer is \( \frac{\mu I}{2\pi r} \).