Physics · Moving Charges and Magnetic Field

NEET (UG) 2023 — Question 42

A very long conducting wire is bent in a semi-circular shape from $A$ to $B$ as shown in figure. The magnetic field at point $P$ for steady current configuration is given by :

Question figure

Option A:
$\frac{\mu_{0} i}{4 R}$ pointed away from the page
Option B:
$\frac{\mu_{0} i}{4 R}\left[1-\frac{2}{\pi}\right]$ pointed away from page
Correct
Option C:
$\frac{\mu_{0} i}{4 R}\left[1-\frac{2}{\pi}\right]$ pointed into the page
Option D:
$\frac{\mu_{0} i}{4 R}$ pointed into the page

Answer: B

Step-by-step solution

$B_P$ due to wire $1=\frac{\mu_0 i}{4 \pi R} \otimes$

$B_P$ due to wire $3=\frac{\mu_0 i}{4 \pi R}$ $B_P$ due to wire $2=\frac{\mu_0 i}{4 R} \odot$

B_{\text {net }}=-\frac{\mu_0 i}{2 \pi R}+\frac{\mu_0 i}{4 R}=\frac{\mu_0 i}{4 R}\left[-\frac{2}{\pi}+1\right]=\frac{\mu_0 i}{4 R}\left[1-\frac{2}{\pi}\right]

Pointed away from page.

Solution figure

Answer key and solution verified before publishing.

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Exam: NEET (UG) 2023
Paper: 2023 paper
Subject: Physics
Chapter: Moving Charges and Magnetic Field
Topic: Magnetic Field Due to Current-Carrying Wire - Biot-Savart Law

A wire carrying a current I along the positive x -axis has length L . It is kept in a magnetic field vecB = (2hati + 3hatj - 4hatk) T . The…