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 \(\mathrm{a} / 2\) and \(2 \mathrm{a}\) is (a) \((1 / 4)\) (b) 4 (c) 1 (d) \((1 / 2)\)

Short answer

The ratio of the magnetic field at \(a/2\) and \(2a\) is 4. The correct answer is (b).

Step-by-step solution

Use Ampère's Law to find the magnetic field expression.

According to Ampère's Law, the magnetic field at a distance \(r\) from the axis of a long straight wire is given by the formula: \[B(r) = \frac{\mu_0 I}{2 \pi r}\] Here, \(B(r)\) is the magnetic field at a distance \(r\), \(\mu_0\) is the permeability of free space, and \(I\) is the current carried by the wire. Now we can use this formula to find the magnetic fields at the distances \(a/2\) and \(2a\).

Find the magnetic field at distance \(a/2\).

By substituting the distance \(a/2\) into the formula from step 1, we can find the magnetic field at this point: \[B\left(\frac{a}{2}\right) = \frac{\mu_0 I}{2 \pi \frac{a}{2}} = \frac{2\mu_0 I}{2 \pi a}\]

Find the magnetic field at distance \(2a\).

By substituting the distance \(2a\) into the formula from step 1, we can find the magnetic field at this point: \[B(2a) = \frac{\mu_0 I}{2 \pi 2a} = \frac{\mu_0 I}{4 \pi a}\]

Calculate the ratio of the magnetic fields.

Now we can compute the ratio of the magnetic fields at the two given distances: \[\frac{B\left(\frac{a}{2}\right)}{B(2a)} = \frac{\frac{2\mu_0 I}{2 \pi a}}{\frac{\mu_0 I}{4 \pi a}} = \frac{1}{1/4} = 4\] So, the ratio of the magnetic field at \(a/2\) and \(2a\) is 4. The correct answer is (b).