Question

A straight conductor carrying current i splits into two parts as shown in the figure. The radius of the circular loop is \( R \). The total magnetic field at the centre \( P \) of the loop is, 

(1) \( \frac{\mu_{0} i}{2 R} \), inward 

(2) Zero

Answer 1

Magnetic field due to i₁ is B₁ = \(\frac{\mu_0i_1}{2R}\frac{\theta_1}{2\pi}\)

Magnetic field due to i₂ is B₂ = \(\frac{\mu_0i_2}{2R}\frac{\theta_2}{2\pi}\)

Parallel combination \(\frac{i_1}{i_2}=\frac{\rho i_2}{A}\times\frac{A}{\rho i_1}=\frac{i_1}{i_2}\)

\(\frac{i_1}{i_2}=\cfrac{\frac142\pi R}{\frac34(2\pi R)}\)

i₁ = \(\frac{i_2}3\)

i₂ = 3i₁

Net magnetic field

 = \(\frac{\mu_0i}{2R}(\frac{\theta_1}{2\pi})-\frac{\mu_0i_2}{2R}(\frac{\theta_2}{2\pi})\) 

\(=\frac{\mu_0}{2R}(\frac{3\pi}{2\times2\pi})-\frac{\mu_0i_2}{2R}(\frac{\pi}{2\times2\pi})\) 

\(=\frac{\mu_0}{2R}[\frac{3i_1}4-\frac{i_2}4]\) 

\(=\frac{\mu_0}{2R}[\frac{3i}4-\frac{3i}4]=0\)