question_answer

A current I flows in a long straight wire whose cross-section is in the shape of a thin quarter ring of radius R. Find the induction of the magnetic field (B) at point O on the axis.

A) \[\frac{\sqrt{2}{{\mu }_{0}}I}{{{\pi }^{2}}R}\]

B) \[\frac{{{\mu }_{0}}I}{{{\pi }^{2}}R}\]

C) \[\frac{\sqrt{3}{{\mu }_{0}}I}{2{{\pi }^{2}}R}\]         

D) \[\frac{3{{\mu }_{0}}I}{\sqrt{2}{{\pi }^{2}}R}\]

Correct Answer: A

Solution :

[a] Consider a long wire element as shown. The figure shows the cross section of the wire. Current through the wire element is \[dI=\frac{I}{\frac{\pi }{2}}d\theta =\frac{2I}{\pi }d\theta \] Magnetic field due to the current dl at O is \[dB=\frac{{{\mu }_{0}}dI}{2\pi R}\] \[=\frac{{{\mu }_{0}}\frac{2I}{\pi }d\theta }{2\pi R}=\frac{{{\mu }_{0}}I}{{{\pi }^{2}}R}d\theta \]  x  component of the field is \[d{{B}_{x}}=\frac{{{\mu }_{0}}I}{{{\pi }^{2}}R}\sin \theta \,\,d\theta \] \[{{B}_{x}}=\int{d{{B}_{x}}}=\frac{{{\mu }_{0}}I}{{{\pi }^{2}}R}\int\limits_{0}^{\pi /2}{\sin \theta d\theta =\frac{{{\mu }_{0}}I}{{{\pi }^{2}}R}}\] From symmetry \[{{B}_{y}}={{B}_{x}}\] \[\therefore \] Resultant B at O is \[B=\sqrt{B_{x}^{2}+B_{y}^{2}}=\frac{\sqrt{2}{{\mu }_{0}}I}{{{\pi }^{2}}R}\]