Question

A current of $10\text{A}$ flows around a closed path in a circuit which is in a horizontal plane as shown in the figure. The circuit consists of eight alternating arcs of radii $r_1=0.08\text{m}$ and $r_2=0.12\text{m}$. Each arc subtends an equal angle at the centre. The magnetic field at the centre will be

• A. $6.5\times {10}^{-5}\text{T}$
• B. $6.5\times {10}^{-6}\text{T}$
• C. $6.5\times {10}^{-7}\text{T}$
• D. $6.5\times {10}^{-4}\text{T}$

Answer 1

The correct option is A $6.5\times {10}^{-5}\text{T}$

As current in all arcs is in the same direction i.e. anticlockwise, so

$B_{\text{net}}=B_{\text{inner arcs}}+B_{\text{outer arcs}}$

As there are an equal number of arcs and each arc subtends the same angle at the centre,

Total length of inner arcs $=\pi r_1$

Total length of outer arcs $=\pi r_2$

$\therefore B_{\text{inner arcs}}=\frac{{\mu }_{o}i}{4\pi r_1^2}\times \pi r_1=\frac{{\mu }_{o}i}{4r_1}$

Similarly,

$\therefore B_{\text{outer arcs}}=\frac{{\mu }_{o}i}{4\pi r_2^2}\times \pi r_2=\frac{{\mu }_{o}i}{4r_2}$

$\Rightarrow B_{\text{net}}=\frac{{\mu }_{o}i}{4}(\frac{1}{r_1}+\frac{1}{r_2})$

Substituting the values,

$B_{net}=\frac{4\pi \times {10}^{-7}\times 10}{4}(\frac{1}{0.08}+\frac{1}{0.12})=6.5\times {10}^{-5}\text{T}$

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