Question

If a straight conductor of length $40\text{cm}$ bent in the form of a square and the current $2\text{A}$ is passing through the square, then the magnetic induction at the center of the square loop is-

• A. $14\sqrt{2}\mu \text{T}$
• B. $16\sqrt{2}\mu \text{T}$
• C. $18\sqrt{2}\mu \text{T}$
• D. $12\sqrt{2}\mu \text{T}$

Answer 1

The correct option is B $16\sqrt{2}\mu \text{T}$



Given, a wire of length $40\text{cm}$ is bent into a square loop.

Now, each side length $l=\frac{L}{4}=\frac{40}{4}=10\text{cm}\text{and}I=2\text{A}$

The net magnetic field at the centre of square loop will be,

$B_{net}=4B⨂$

Where, $B$ is field due to a side.

$=4\times \frac{{\mu }_{0}i}{4\pi r}(\sin {45}^{\circ }+\sin {45}^{\circ })$

$=\frac{4\times {10}^{-7}\times 2\times \sqrt{2}}{5\times {10}^{-2}}$

$=16\sqrt{2}\times {10}^{-6}\text{T}\text{(or)}$

$=16\sqrt{2}\mu \text{T}$

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