A wire carrying current $\text{I}$ has the shape as shown in adjoining figure. Linear parts of the wire are very long and parallel to $\text{X}$ -axis while semicircular portion of radius $\text{R}$ is lying in $\text{Y}$ -$\text{Z}$ plane. Magnetic field at point $\text{O}$ is : ![NEET 2015 Physics, Magnetic Effects of Current — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/616059730283de43c87e3e22/Physics/images/Magnetic_Effects_of_Current/66bcbf3aae81d148616b1f9f/question_1__q_66bcbf3aae81d148616b1f9f__cdn-question-pool.getmarks.app__image16-adecdbfe-2521-417b-ab71-53b218517e47__e2d3a5de35_final_ppt_sync.jpg)

![NEET 2015 Physics, Magnetic Effects of Current — solution figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/616059730283de43c87e3e22/Physics/images/Magnetic_Effects_of_Current/66bcbf3aae81d148616b1f9f/solution_1__q_66bcbf3aae81d148616b1f9f__cdn-question-pool.getmarks.app__image17-8685bcdb-66ce-47d9-9bdc-f86b33591875__6b83b1b9ad_final_ppt_sync.png) ${\vec{B}}_{c} = {\vec{B}}_{1}+{\vec{B}}_{2}+{\vec{B}}_{3}.$ ${B}_{1} = \frac{{\mu }_{0}}{4\pi R} (\mathrm{sin}⁡90+sin\theta )-\hat{k}$ ${B}_{1} = \frac{{\mu }_{0}}{4\pi R} (-\hat{k})={B}_{3}$ B due to segment 2 ${B}_{2} = \frac{{\mu }_{0} I}{4\pi }\times \pi (-\hat{i})=-\frac{{\mu }_{0}I}{4R}\hat{i}$ So B at center ${\vec{B}}_{c}= {\vec{B}}_{1}+ {\vec{B}}_{2}+{\vec{B}}_{3}$ ${B}_{c}= \frac{-{\mu }_{0}I}{4R} (i+\frac{2\hat{k}}{\pi })= \frac{-{\mu }_{0}I}{4\pi R} (\pi \hat{i}+2\hat{k})$

NEET UG 2015 — Physics Electromagnetism

2015mcqhard

A wire carrying current $\text{I}$ has the shape as shown in adjoining figure. Linear parts of the wire are very long and parallel to $\text{X}$ -axis while semicircular portion of radius $\text{R}$ is lying in $\text{Y}$ - $\text{Z}$ plane. Magnetic field at point $\text{O}$ is :

NEET 2015 Physics, Magnetic Effects of Current — question figure

Official previous-year question

Held on 30 Apr 2015 · Verified 9 Jul 2026.

Options

A
$\vec{B}=\frac{{\mu }_{0}}{4\pi }\frac{I}{R} (\pi \hat{i}+2\hat{k})$
B
$\vec{B}=-\frac{{\mu }_{0}}{4\pi }\frac{I}{R} (\pi \hat{i}-2\hat{k})$
C
$\vec{B}=-\frac{{\mu }_{0}}{4\pi }\frac{I}{R} (\pi \hat{i}+2\hat{k})$
D
$\vec{B}=\frac{{\mu }_{0}}{4\pi }\frac{I}{R} (\pi \hat{i}-2\hat{k})$

Did you get this right?

Solution

NEET 2015 Physics, Magnetic Effects of Current — solution figure

${\vec{B}}_{c} = {\vec{B}}_{1}+{\vec{B}}_{2}+{\vec{B}}_{3}.$

${B}_{1} = \frac{{\mu }_{0}}{4\pi R} (\mathrm{sin}⁡90+sin\theta )-\hat{k}$

${B}_{1} = \frac{{\mu }_{0}}{4\pi R} (-\hat{k})={B}_{3}$

B due to segment 2

${B}_{2} = \frac{{\mu }_{0} I}{4\pi }\times \pi (-\hat{i})=-\frac{{\mu }_{0}I}{4R}\hat{i}$

So B at center ${\vec{B}}_{c}= {\vec{B}}_{1}+ {\vec{B}}_{2}+{\vec{B}}_{3}$

${B}_{c}= \frac{-{\mu }_{0}I}{4R} (i+\frac{2\hat{k}}{\pi })= \frac{-{\mu }_{0}I}{4\pi R} (\pi \hat{i}+2\hat{k})$

Your note

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Options

$\vec{B}=\frac{{\mu }_{0}}{4\pi }\frac{I}{R} (\pi \hat{i}+2\hat{k})$

$\vec{B}=-\frac{{\mu }_{0}}{4\pi }\frac{I}{R} (\pi \hat{i}-2\hat{k})$

$\vec{B}=-\frac{{\mu }_{0}}{4\pi }\frac{I}{R} (\pi \hat{i}+2\hat{k})$

$\vec{B}=\frac{{\mu }_{0}}{4\pi }\frac{I}{R} (\pi \hat{i}-2\hat{k})$

Solution

NEET 2015 Physics, Magnetic Effects of Current — solution figure

${\vec{B}}_{c} = {\vec{B}}_{1}+{\vec{B}}_{2}+{\vec{B}}_{3}.$

${B}_{1} = \frac{{\mu }_{0}}{4\pi R} (\mathrm{sin}⁡90+sin\theta )-\hat{k}$

${B}_{1} = \frac{{\mu }_{0}}{4\pi R} (-\hat{k})={B}_{3}$

B due to segment 2

${B}_{2} = \frac{{\mu }_{0} I}{4\pi }\times \pi (-\hat{i})=-\frac{{\mu }_{0}I}{4R}\hat{i}$

So B at center ${\vec{B}}_{c}= {\vec{B}}_{1}+ {\vec{B}}_{2}+{\vec{B}}_{3}$

${B}_{c}= \frac{-{\mu }_{0}I}{4R} (i+\frac{2\hat{k}}{\pi })= \frac{-{\mu }_{0}I}{4\pi R} (\pi \hat{i}+2\hat{k})$

NEET UG 2015 — Physics Electromagnetism

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Solution

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NEET UG 2015 — Physics Electromagnetism

Options

Did you get this right?

Solution

Your note

NEET UG Electromagnetism in other years

More NEET UG Electromagnetism questions

Other Physics topics for NEET UG

Mechanics

Modern Physics

Thermodynamics

Optics

Waves & Oscillations

Keep practicing Electromagnetism