Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Medical
  • Two identical long conducting wires AOB and COD are placed at right angle to each other, with one above other such that ' O ' is their common point for the two. The wires carry <i

Two identical long conducting wires AOB and COD are placed at right angle to each other, with one above other such that ' O ' is their common point for the two. The wires carry \mathrm{I_1}  and \mathrm{I_2}  currents respectively. Point ' P ' is lying at distance ' d ' from ' O ' along a direction perpendicular to the plane containing the wires. The magnetic field at the point ' P ' will be :

Option: 1

\mathrm{\frac{\mu_0}{2 \pi d}\left(\frac{I_1}{I_2}\right)}


Option: 2

\mathrm{\frac{\mu_0}{2 \pi \mathrm{d}}\left(\mathrm{I}_1+\mathrm{I}_2\right)}


Option: 3

\mathrm{\frac{\mu_0}{2 \pi d}\left(I_1^2-I_2^2\right)}


Option: 4

\mathrm{\frac{\mu_0}{2 \pi r}\left(I_1^2 \times I_2^2\right)^{\frac{1}{2}}}


Answers (1)

best_answer

Net magnetic field,    \mathrm{B}=\sqrt{\mathrm{B}_1^2+\mathrm{B}_2^2}

\mathrm{ =\sqrt{\left(\frac{\mu_0 I_1}{2 \pi d}\right)^2+\left(\frac{\mu_0 I_2}{2 \pi d}\right)^2} \\ }

\mathrm{ \left(B_1=\frac{\mu_0 1_1}{2 \pi d} \text { and } B_2=\frac{\mu_0 1_2}{2 \pi d}\right) \\ }

\mathrm{ =\frac{\mu_0}{2 \pi d} \sqrt{I_1^2+I_2^2} }.
 

Posted by

avinash.dongre

View full answer

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks

Similar Questions

Trending Articles/News

anam-khan

Before NEET, CMC Vellore’s unique MBBS admissions tested aptitude along with merit; paper-leak restarts debate
anam-khan

NEET-UG paper leak fallout: Latur admin orders crackdown on illegal coaching classes
anam-khan

Re-NEET in 30 days a ‘massive logistical challenge’, NTA adopts ‘zero trust’ policy: Report

Latest Question