Question

In the given diagram, a single infinitely long, straight wire has been bent into an arc of a circle of radius $50\text{mm}$ and two long straight portions. Find the magnetic field at the origin $O$, when a current of $5\text{A}$ flows through the wire.

• A. $4.7\times {10}^{-5}\text{T}$
• B. $9.2\times {10}^{-5}\text{T}$
• C. $7.2\times {10}^{-5}\text{T}$
• D. $3.7\times {10}^{-5}\text{T}$

Answer 1

The correct option is D $3.7\times {10}^{-5}\text{T}$

Segments $1$ and $3$ are two semi-infinite long straight wires.

Therefore, the magnetic field produced by them will have the same magnitude. i.e,

${\overrightarrow{B}}_1={\overrightarrow{B}}_3=\frac{{\mu }_{o}I}{4\pi r}(-\hat{k})$

Now, magnetic field due to segment $2$,

${\overrightarrow{B}}_2=\frac{{\mu }_{o}I}{4r}(-\hat{i})$

Net magnetic field,

${\overrightarrow{B}}_{net}=2\times \frac{{\mu }_{o}I}{4\pi r}(-\hat{k})+\frac{{\mu }_{o}I}{4r}(-\hat{i})$

$\therefore B_{net}=\sqrt{{\left(\frac{{\mu }_{o}I}{2\pi r}\right)}^2+{\left(\frac{{\mu }_{o}I}{4r}\right)}^2}$

$=\frac{{\mu }_{o}I}{4\pi r}\sqrt{{\pi }^2+4}$

$=\frac{{10}^{-7}\times 5}{50\times {10}^{-3}}\sqrt{{\pi }^2+4}$

$=3.7\times {10}^{-5}\text{T}$

Hence, option $\left(d\right)$ is the correct answer.