Question

Figure shows current loop having two circular arcs joined by two radial lines. Find the magnetic field $B$ at the center $\text{O}$. [Take $R_1=5\text{cm},R_2=10\text{cm},I=\frac{2}{\pi }\text{A}$ and $\varphi =\frac{\pi }{4}\text{rad}$]

• A. $0.05\mu \text{T}$
• B. $0.5\mu \text{T}$
• C. $5\mu \text{T}$
• D. Zero

Answer 1

The correct option is B $0.5\mu \text{T}$

We know due to complete circle at the center is,

$B=\frac{{\mu }_{0}I}{2R}$

For a circular segment (or) arc at the center is,


$B=\frac{{\mu }_{0}I}{2R}\times \frac{\varphi }{2\pi }$

Considering the arc$\text{AB}$, field at the center of the circular arc is,

$B_{AB}=\frac{{\mu }_{0}I}{2R_2}\times \frac{\varphi }{2\pi }$ ( Into the plane)

$B_{BC}=B_{AD}=0$ (Field point lies on the line of current element)

Considering the arc$\text{CD}$, field at the center of the circular arc is,

$B_{CD}=\frac{{\mu }_{0}I}{2R_1}\times \frac{\varphi }{2\pi }$ ( Out to the plane)

$B_{net}=B_{in}-B_{out}$

$=\frac{{\mu }_{0}I}{2R_2}\times \frac{\varphi }{2\pi }-\frac{{\mu }_{0}I}{2R_1}\times \frac{\varphi }{2\pi }$

$=\frac{{\mu }_{0}I\varphi }{4\pi }(\frac{1}{R_2}-\frac{1}{R_1})$

$=\frac{{\mu }_{0}I\varphi }{4\pi \left(R_1{R}_2\right)}(R_2-R_1)$

Substituting the data given in the question,

$B_{net}=\frac{4\pi \times {10}^{-7}\times \frac{\pi }{4}\times \frac{2}{\pi }\times (10-5)}{4\pi \times 10\times 5}=0.5\mu textT$

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