A loop ABCDA , carrying current I=12 ~A , is placed in a plane, consists of two semi-circular segments of radius R_1=6 π ~m and R_2=4 π ~m . The magnitude of the resultant magnetic field at center O is k × 10^ -7 ~T .

Correct answer: 0 Magnetic field due to AB \& CD=0 aligned & B_0=|B_ R_1 -B_ R_2 | & = μ_0 I 4 R_2 - μ_0 I 4 R_1 & = 4 π × 10^ -7 × 12 4 ((1)/(4 π)-(1)/(6 π)) & =12 π × 10^ -7 ((1)/(12 π)) & =1 × 10^ -7 aligned K=1

JEE Main 20253 AprilMorning Shift

Exam: JEE Main
Year: 2025
Date: 3 April
Shift: Morning Shift
Subject: Physics
Chapter: Magnetic Effects of Current

JEE Main 2025 · 3 April · Morning Shift · Physics · Magnetic Effects of Current

A loop ABCDA , carrying current I=12 ~A , is placed in a plane,...

A loop ABCDA , carrying current

\mathrm{I}=12 \mathrm{~A}

, is placed in a plane, consists of two semi-circular segments of radius

R_1=6 \pi \mathrm{~m}

and

\mathrm{R}_2=4 \pi \mathrm{~m}

. The magnitude of the resultant magnetic field at center O is

\mathrm{k} \times 10^{-7} \mathrm{~T}

. The value of k is ______ (Given

\mu_0=4 \pi \times 10^{-7} \mathrm{Tm} \mathrm{A}^{-1}

)

A loop ABCDA , carrying current I=12 ~A , is placed in a plane, consists of two semi-circular segments of radius R_1=6

Correct answer

Step-by-step solution

Magnetic field due to

\mathrm{AB} \& \mathrm{CD}=0

\begin{aligned}<br/>& \mathrm{B}_0=\left|\mathrm{B}_{\mathrm{R}_1}-\mathrm{B}_{\mathrm{R}_2}\right| \\ & =\frac{\mu_0 \mathrm{I}}{4 \mathrm{R}_2}-\frac{\mu_0 \mathrm{I}}{4 \mathrm{R}_1} \\ & =\frac{4 \pi \times 10^{-7} \times 12}{4}\left(\frac{1}{4 \pi}-\frac{1}{6 \pi}\right) \\ & =12 \pi \times 10^{-7}\left(\frac{1}{12 \pi}\right) \\ & =1 \times 10^{-7}<br/>\end{aligned}

K=1

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