Question

The magnetic field at the point $O$ is

• A. $1·5x{10}^{-5}$
• B. ${10}^{-5}$
• C. $2x{10}^{-5}$
• D. ${10}^{-4}$

Answer 1

The correct option is B

${10}^{-5}$

Step 1. Given data:

From the figure, we get the angle made at the center$O={45}^{\circ }$.

The radius of the smaller segment$=3cm$.

The radius of the smaller segment$=5cm$.

The current$\left(I\right)=10A$

Step 2. Formula used:

The magnetic field at the center due to the arc is,

$B=\frac{{\mu }_{0}I}{4\mathrm{\pi r}}\theta$ (${\mu }_0=$Free space permeability, $r=$The radius of arc)

Step 3. Calculating the magnetic field at the center

Let us break this into four segments $AB,BD,CD,AC$.

Of these four only $AB,CD$ segments will contribute because the other two segments are straight lines and the point lies along the straight lines.

The magnetic field due to the straight segments will be zero.

Due to the arc segment $CD$, the magnetic field is,

$\begin{array}{rcl}{B}_1& =& \frac{{\mu }_{0}I}{4\mathrm{\pi }\times 0.03}\times \frac{\mathrm{\pi }}{4}\left(\theta ={45}^{\circ }=\frac{\mathrm{\pi }}{4}\right)\\ & =& \frac{4\mathrm{\pi }\times {10}^{-7}\times 10}{4\mathrm{\pi }\times 0.03}\times \frac{\mathrm{\pi }}{4}\\ & =& \frac{\mathrm{\pi }}{4}\times 3.33\times {10}^{-5}T\end{array}$

Due to the arc segment$AB$, the magnetic field is,

$\begin{array}{rcl}{B}_1& =& \frac{{\mu }_{0}I}{4\mathrm{\pi }\times 0.05}\times \frac{\mathrm{\pi }}{4}\left(\theta ={45}^{\circ }=\frac{\mathrm{\pi }}{4}\right)\\ & =& \frac{4\mathrm{\pi }\times {10}^{-7}\times 10}{4\mathrm{\pi }\times 0.05}\times \frac{\mathrm{\pi }}{4}\\ & =& \frac{\mathrm{\pi }}{4}\times 2\times {10}^{-5}T\end{array}$

Then the resultant magnetic field for the ring portion is,

$\begin{array}{rcl}{B}_1-B_2& =& \left(3.33-2\right)\times {10}^{-5}\times \left(\frac{\mathrm{\pi }}{4}\right)\\ & =& {10}^{-5T}\end{array}$

Thus, the resultant magnetic field is${10}^{-5}T$

Hence, option B is the correct answer.