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 \(\frac{a}{2}\) and \(2a\) from axis of the wire is:
- 1 : 4
- 4 : 1
- 1 : 1
- 3 : 4
The Correct Option is C
Approach Solution - 1
To determine the ratio of the magnetic field at distances \(\frac{a}{2}\) and \(2a\) from the axis of a long straight wire carrying a steady current \(I\), we use Ampere's Law. This law relates the integrated magnetic field around a closed loop to the electric current passing through the loop.
Ampere's Law is given by:
\(\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{\text{enclosed}}\)
Where \(\mathbf{B}\) is the magnetic field, \(d\mathbf{l}\) is the differential length element of the loop, \(\mu_0\) is the permeability of free space, and \(I_{\text{enclosed}}\) is the current enclosed by the loop.
- Inside the wire (at \(r = \frac{a}{2}\)):
- Current density, \(J = \frac{I}{\pi a^2}\), since the current is uniformly distributed.
- Current enclosed, \(I_{\text{enclosed}} = J \times \pi \left(\frac{a}{2}\right)^2 = \frac{I}{\pi a^2} \times \frac{\pi a^2}{4} = \frac{I}{4}\).
- Applying Ampere's Law: \(\oint \mathbf{B} \cdot d\mathbf{l} = B\left(2\pi \frac{a}{2}\right) = \mu_0 \frac{I}{4}\).
- Simplifying gives the magnetic field, \(B_{\frac{a}{2}} = \frac{\mu_0 I}{4\pi a}\).
- Outside the wire (at \(r = 2a\)):
- The entire current \(I\) is enclosed since \(r = 2a\) is outside the wire.
- Applying Ampere's Law: \(\oint \mathbf{B} \cdot d\mathbf{l} = B(2\pi \times 2a) = \mu_0 I\).
- Simplifying gives the magnetic field, \(B_{2a} = \frac{\mu_0 I}{4\pi a}\).
Now, calculate the ratio of magnetic fields:
\(\frac{B_{\frac{a}{2}}}{B_{2a}} = \frac{\frac{\mu_0 I}{4\pi a}}{\frac{\mu_0 I}{4\pi a}} = 1 : 1\)
The ratio of the magnetic field at \(\frac{a}{2}\) to that at \(2a\) is therefore 1 : 1.
Approach Solution -2
For a point inside the wire (\(r<a\)), the magnetic field is given by:
\[B_1 \times 2\pi \frac{a}{2} = \mu_0 \frac{I}{4} \implies B_1 = \frac{\mu_0 I}{4\pi a}.\]
For a point outside the wire (\(r>a\)), the magnetic field is:
\[B_2 \times 2\pi \times 2a = \mu_0 I \implies B_2 = \frac{\mu_0 I}{4\pi a}.\]
The ratio of magnetic fields:
\[\frac{B_1}{B_2} = \frac{\frac{\mu_0 I}{4\pi a}}{\frac{\mu_0 I}{4\pi a}} = 1 : 1.\]
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