A wire carrying current I is bent in the shape ABCDEFA as shown, where rectangles ABCDA and ADEFA are perpendicular to each other. If the sides of the rectangles are of lengths a and b, then the magnitude and direction of magnetic moment of the loop ABCDEFA is:
A wire carrying current I is bent in the shape ABCDEFA as shown, where rectangles ABCDA and ADEFA are perpendicular to each other. If the sides of the rectangles are of lengths a and b, then the magnitude and direction of magnetic moment of the loop ABCDEFA is:
Option 1 - <p>√2 abI, along (ĵ/√2 + k̂/√2)<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--></p>
Option 2 - <p>√2 abI, along (ĵ/√5 + 2k̂/√5)</p>
Option 3 - <p>abI, along (ĵ/√2 + k̂/√2)</p>
Option 4 - <p>abI, along (ĵ/√5 + k̂/√5)</p>
7 Views|Posted 5 months ago
Asked by Shiksha User
1 Answer
A
Answered by
5 months ago
Correct Option - 1
Detailed Solution:
m = I (abk + abj)
|m| = Iab√2
Direction ⇒ (j+k)/√2
Similar Questions for you
Ohm's law is valid if I depends on V' linearly.
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else.
On Shiksha, get access to
66K
Colleges
|
1.2K
Exams
|
6.9L
Reviews
|
1.8M
Answers
Learn more about...

Physics Current Electricity 2025
View Exam DetailsMost viewed information
SummaryDidn't find the answer you were looking for?
Search from Shiksha's 1 lakh+ Topics
or
Ask Current Students, Alumni & our Experts
Have a question related to your career & education?
or
See what others like you are asking & answering





