Question

What is the magnetic field at point $\text{P}$ if $\text{AP}$ is making an angle ${15}^{\circ }$ as shown in the figure.

• A. $\frac{{\mu }_0}{4\pi }\frac{I}{2a}[\sqrt{3}-\sqrt{2}]$
• B. $\frac{{\mu }_0}{2\pi }\frac{I}{a}[\sqrt{3}-\sqrt{2}]$
• C. $\frac{{\mu }_0}{\pi }\frac{I}{a}[\sqrt{3}-\sqrt{2}]$
• D. none of these

Answer 1

The correct option is A $\frac{{\mu }_0}{4\pi }\frac{I}{2a}[\sqrt{3}-\sqrt{2}]$



From $△OAP$,

We can see $OP=OA=4\text{m}$

$\angle POA={90}^{\circ }$

So it is a isosceles right angled triangle

$\angle OAP=\angle APO={45}^{\circ }$

Given, $\angle APB={15}^{\circ }$

$\therefore \angle OPB={60}^{\circ }$

Now magnetic field due to $\text{AB}$ at point $\text{P}$ is

$B=\frac{{\mu }_0}{4\pi }\frac{I}{a}[\sin {60}^{\circ }-\sin {45}^{\circ }]$

$=\frac{{\mu }_0}{4\pi }\frac{I}{a}[\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}]$

$=\frac{{\mu }_0}{4\pi }\frac{I}{2a}[\sqrt{3}-\sqrt{2}]$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option (a) is the correct answer.