Question

A current loop consists of two identical semicircular parts each of radius \(2R\), one lying in the \(x-y\) plane and the other in \(x-z\) plane. If the current in the loop is \(I\), the resultant magnetic field due to two semicircular parts at their common centre is

• A. \( \frac{\mu_0 I}{4R} \)
• B. \( \frac{\mu_0 I}{2\sqrt{2}R} \)
• C. \( \frac{\mu_0 I}{\sqrt{2}R} \)
• D. \( \frac{\mu_0 I}{4\sqrt{2}R} \)

Hint:

For a semicircular wire, \(B = \frac{\mu_0 I}{4r}\). If two equal magnetic fields are perpendicular, resultant is \(B\sqrt{2}\).

Answer 1

The Correct Option is D

Step 1: Magnetic field due to a circular arc.
For a semicircular current-carrying wire of radius \(r\), magnetic field at centre is:
\[ B = \frac{\mu_0 I}{4r} \]

Step 2: Substitute radius of each semicircle.
Here, radius of each semicircle is \(2R\).
\[ B = \frac{\mu_0 I}{4(2R)} \]
\[ B = \frac{\mu_0 I}{8R} \]

Step 3: Direction of magnetic fields.
One semicircle lies in the \(x-y\) plane, so its magnetic field is along the \(z\)-axis. The other semicircle lies in the \(x-z\) plane, so its magnetic field is along the \(y\)-axis.

Step 4: Angle between two magnetic fields.
The two magnetic fields are perpendicular to each other.
\[ \theta = 90^\circ \]

Step 5: Resultant magnetic field.
\[ B_{\text{net}} = \sqrt{B^2 + B^2} \]
\[ B_{\text{net}} = \sqrt{2}B \]

Step 6: Substitute value of \(B\).
\[ B_{\text{net}} = \sqrt{2} \times \frac{\mu_0 I}{8R} \]
\[ B_{\text{net}} = \frac{\mu_0 I}{4\sqrt{2}R} \]
\[ \boxed{\frac{\mu_0 I}{4\sqrt{2}R}} \] Hence, correct answer is option (D).