Question

A very long wire $\text{ABDMNDC}$ is shown in figure carrying current $I.\text{AB and BC}$ parts are straight, long and at right angle. At $D$ wire forms a circular turn $\text{DMND}$ of radius $R$. $\text{AB, BC}$ parts are tangential to circular turn at $N\text{and}D.$ Magnetic field at the centre of circle is

• A. $\frac{{\mu }_{0}I}{2\pi R}(\pi +\frac{1}{\sqrt{2}})$
• B. $\frac{{\mu }_{0}I}{2\pi R}(\pi -\frac{1}{\sqrt{2}})$
• C. $\frac{{\mu }_{0}I}{2\pi R}(\pi +1)$
• D. $\frac{{\mu }_{0}I}{2R}$

Answer 1

The correct option is A $\frac{{\mu }_{0}I}{2\pi R}(\pi +\frac{1}{\sqrt{2}})$

Magnetic field due to AB is $B_1=\frac{{\mu }_{0}I}{4\pi R}[\sin {90}^{\circ }-\sin {45}^{\circ }]⨂$
Magnetic field due to BD is $B_2=\frac{{\mu }_{0}I}{4\pi R}[\sin {45}^{\circ }+\sin {90}^{\circ }]⨀$
Magnetic field due to circular coil is$B_3=\frac{{\mu }_{0}I}{2R}⨀$
Net Magnetic field at center is
$B_0=B_1+B_2+B_3$

$B_0=\frac{{\mu }_{0}I}{4\pi R}[\sin {90}^{\circ }-\sin {45}^{\circ }]⨂+\frac{{\mu }_{0}I}{2R}⨀+\frac{{\mu }_{0}I}{4\pi R}[\sin {45}^{\circ }+\sin {90}^{\circ }]⨀$

$B_0=(-\frac{{\mu }_{0}I}{4\pi R}(1-\frac{1}{\sqrt{2}})+\frac{{\mu }_{0}I}{2R}+\frac{{\mu }_{0}I}{4\pi R}(1+\frac{1}{\sqrt{2}}))⨀$

$B_0=\frac{{\mu }_{0}I}{2\pi R}(\pi +\frac{1}{\sqrt{2}})⨀$