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 (3) $3 \mu_{0} / 32 R$, outward (4) $3 \mu_{0} i / 32 R$, inward

Home » 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) , inward (2) Zero (3) , outward (4) , inward

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 (3) $3 \mu_{0} / 32 R$ , outward (4) $3 \mu_{0} i / 32 R$ , inward

Physics | Physics

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 (3) $3 \mu_{0} / 32 R$ , outward (4) $3 \mu_{0} i / 32 R$ , inward

Answer (2)

Solution:

Net magnetic field at point ‘P’
$\mathbf{B}_{\text {net }}=\overrightarrow{\mathbf{B}_{1}}+\overline{\mathbf{B}_{2}}$
Here $\bar{B}_{1}$ and $\bar{B}_{2}$ are equal in magnitude and opposite in direction.
Hence, $\mathbf{B}_{\text {net }}=\mathbf{B}_{1}-\mathbf{B}_{2}$
$\vec{i}_{1}=i\left(\frac{\theta}{2 \pi}\right) \Rightarrow \mathbf{B}_{1}=\frac{\mu_{0} \dot{L}_{1}}{2 R}\left(\frac{2 \pi-\theta}{2 \pi}\right)$
$\dot{i}_{2}=i\left(\frac{2 \pi-\theta}{2 \pi}\right) \Rightarrow \mathbf{B}_{2}=\frac{\mu_{0} \mathrm{i}_{2}}{2 R}\left(\frac{\theta}{2 \pi}\right)$
$B_{\text {net }}=B_{1}-B_{2}=0$