Question

A current carrying wire is in form of square loop of side length $10cm.$ Current in wire is $5A.$ Find out magnetic field at center of loop.

• A. ${10}^{-7}T$
• B. $\sqrt{2}\times {10}^{-7}T$
• C. $4\sqrt{2}\times {10}^{-5}T$
• D. $4\times {10}^{-7}T$

Answer 1

Answer: (C)

$4\sqrt{2}\times {10}^{-5}T$

Given: a=10 cm ,i=5A

Solution: Formula used:

$B=\frac{{\mu }_{0}i}{4\pi r}(sin\alpha +sin\beta )$

Rember here that r=$\frac{a}{2}$ (because centre is at half the distance of total )

Now put values, remember $\alpha ,\beta ={45}^0$

After solving $B=\frac{{10}^{-7}\times 5}{5\times {10}^{-2}}(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}})$

$B=\sqrt{2}\times {10}^{-5}$

But ,this is only for one side

for 4 sides $B=4\times \sqrt{2}\times {10}^{-5}$

Hence the correct option is C