Question

An electric current is flowing in a very long pin as shown in the figure. The value of magnetic flux density at point $\text{O}$ will be-

• A. $\frac{{\mu }_{0}i}{4\pi R}[\pi +2]$
• B. $\frac{{\mu }_{0}i}{4\pi R}[\pi +1]$
• C. $\frac{{\mu }_{0}i}{4\pi R}[\pi -2]$
• D. $\frac{{\mu }_{0}i}{4\pi R}[\pi -1]$

Answer 1

The correct option is A $\frac{{\mu }_{0}i}{4\pi R}[\pi +2]$



From, the figure segment $\text{CD}$ and $\text{FG}$ are semi -infinite wires and segment $\text{DEF}$ is a semicircle.

The net magnetic field point $\text{O}$ is,

$B_o=B_{CD}+B_{DEF}+B_{FG}......\left(1\right)$

$B_{CD}=B_{FG}=\frac{{\mu }_{0}i}{4\pi R}⨀$

$B_{DEF}=\frac{{\mu }_{0}i}{2R}\left(\frac{\theta }{2\pi }\right)=\frac{{\mu }_{0}i}{4R}⨀[\because \theta =\pi ]$

From (1) we get,

$\therefore B_0=\frac{{\mu }_{0}i}{4\pi R}+\frac{{\mu }_{0}i}{4\pi R}+\frac{{\mu }_{0}i}{4R}$

$B_0=\frac{{\mu }_{0}i}{4R}[\frac{1}{\pi }+\frac{1}{\pi }+1]=\frac{{\mu }_{0}i}{4R}[\frac{2}{\pi }+1]$

$B_0=\frac{{\mu }_{0}i}{4\pi R}[2+\pi ]$

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