Question

Magnetic field at P due to given structure is

• A. $\left(\frac{{\mu }_{\circ }}{4\pi }\right)\frac{I\theta }{2R}$
• B. $\frac{{\mu }_{\circ }}{4\pi }\frac{2I\theta }{5R}$
• C. $\left(\frac{{\mu }_{\circ }}{4\pi }\right)\frac{5I\theta }{6R}$
• D. $\left(\frac{{\mu }_{\circ }}{4\pi }\right)\frac{2I\theta }{6R}$

Answer 1

The correct option is C $\left(\frac{{\mu }_{\circ }}{4\pi }\right)\frac{5I\theta }{6R}$

Given: Current=i

Solution:Magnetic field due to the structure at P is given by:

$\overrightarrow{B_p}=\overrightarrow{B_1}+\overrightarrow{B_2}+\overrightarrow{B_3}+\overrightarrow{B_4}+\overrightarrow{B_5}$

But, $\overrightarrow{B_2},\overrightarrow{B_4}=0$( angle subtend by straight current carrying conductor is zero)

$\overrightarrow{B_p}=\overrightarrow{B_1}+\overrightarrow{B_3}+\overrightarrow{B_5}$

$\overrightarrow{B_1}$ directed downwards according to right hand thumb rule

$\overrightarrow{B_3}$directed upwards according to right hand thumb rule

$\overrightarrow{B_5}$directed downwards according to right hand thumb rule

So,$\overrightarrow{B_p}=-\overrightarrow{B_1}+\overrightarrow{B_3}-\overrightarrow{B_5}$

B due to a circular current carrying conductor at it's centre is given by:

$B=\frac{{\mu }_{0}i}{4\pi r}.\theta$

$So,\overrightarrow{B_p}=-\frac{{\mu }_{0}i}{4\pi R}.\theta +\frac{{\mu }_{0}i}{4\pi 2R}.\theta -\frac{{\mu }_{0}i}{4\pi 3R}.\theta$

$|\overrightarrow{B_p}|=|-\frac{{\mu }_{0}i}{4\pi R}.\theta [1-\frac{1}{2}+\frac{1}{3}]|$

$|\overrightarrow{B_p}|=\frac{{\mu }_{0}i}{4\pi R}.\frac{5}{6}$

$|\overrightarrow{B_p}|=\left(\frac{{\mu }_{0}i}{4\pi }\right).\frac{5i\theta }{6R}$

Hence the correct answer is C