Question

A regular polygon of $4$ sides is formed by bending wires of total length $1\text{m}$ which carries a current of $10\text{A}$. If the same polygon is reshaped into a circle with the same center then find the change in magnetic field at the center of the polygon ${}^{\prime }{O}^{\prime }$.

• A. $5.25\text{T}$
• B. $5.25\mu \text{T}$
• C. $5.25\text{nT}$
• D. $5.25\text{mT}$

Answer 1

The correct option is B $5.25\mu \text{T}$

Magnetic field of an $n$ sided polygon is given by $$\overrightarrow{B_1}=\frac{{\mu }_{0}I{n}^2\sin \left(\frac{\pi }{n}\right)\tan \left(\frac{\pi }{n}\right)}{2{\pi }^{2}r}⨀$$From the data given in the question, $n=4,r=\frac{1}{2\pi }\text{m},I=10\text{A}$

$\Rightarrow B_1=\frac{4\pi \times {10}^{-7}\times 10\times (4{)}^2\times \sin (\frac{\pi }{4})\tan (\frac{\pi }{4})}{2\times {\pi }^2\times \frac{1}{2\pi }}$

$\Rightarrow B_1=640\times {10}^{-7}\times \sin (\frac{\pi }{4})\tan (\frac{\pi }{4})$

$\Rightarrow B_1=\frac{64}{\sqrt{2}}\mu \text{T}\text{or}45.25\mu \text{T}$

Now the wire is modified into a circle, circle has infinite number of sides. So,

Magnetic field at the center of circle is $$\overrightarrow{B_2}=\frac{{\mu }_{0}I}{2r}$$Thus, $B_2=\frac{4\pi \times {10}^{-7}\times 10}{2\times \frac{1}{2\pi }}$

$\Rightarrow B_2=4{\pi }^2\mu \text{T}\text{or}40\mu \text{T}$

Change in magnetic field at the center be $B=|B_1-B_2|=5.25\mu \text{T}$

Hence, option (b) is the correct answer.