Moving Charges and Magnetism - Result Question 31

32. 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,

[NEET Odisha 2019]

(a) $\frac{\mu_0 i}{2 R}$, inward

(b) Zero

(c) $3 \mu_0 i / 32 R$, outward

(d) $3 \mu_0 i / 32 R$, inward

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Answer:

Correct Answer: 32. (b)

Solution:

  1. (b) Net magnetic field at point ’ $P$ '

$B _{\text{net }}= \vec{B} _1+ \vec{B} _2$

Hence, $B _{\text{net }}=B_1-B_2$

$i_1=i(\frac{\theta}{2 \pi}) \Rightarrow B_1=\frac{\mu_0 i_1}{2 R}(\frac{2 \pi-\theta}{2 \pi})$

$i_2=i(\frac{2 \pi-\theta}{2 \pi}) \Rightarrow B_2=\frac{\mu_0 i_2}{2 R}(\frac{\theta}{2 \pi})$

Here $\theta=\frac{\pi}{2}$

$B _{\text{net }}=B_1-B_2=0$