Question

If the magnetic field at 'P' in the given figure can be written as K tan $\left(\frac{a}{2}\right)$ then k is

• A. $\frac{{\mu }_{0}I}{4\pi d}$
• B. $\frac{{\mu }_{0}I}{2\pi d}$
• C. $\frac{{\mu }_{0}I}{\pi d}$
• D. $\frac{{2\mu }_{0}I}{\pi d}$

Answer 1

The correct option is B $\frac{{\mu }_{0}I}{2\pi d}$

$Solution:$

Let us compute the magnetic field due to any one segment:

$B=\frac{{\mu }_{0}i}{4\pi \left(dsin\alpha \right)}(cos{0}^o+cos(180-\alpha \left)\right)$

=$\frac{{\mu }_{0}I}{4\pi \left(dsin\alpha \right)}(1-cos\alpha )$

=$\frac{{\mu }_{0}I}{4\pi d}\frac{tan\left(\alpha \right)}{2}$

Hence, resultant field will be

$B_{net}=2B=\frac{{\mu }_{0}I}{2\pi d}\frac{tan\left(\alpha \right)}{2}$

$⟹K=\frac{{\mu }_{0}I}{2\pi d}$

$\text{Hence B is the correct option}$