wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Two identical short bar magnets of magnetic moment M and 3M are placed at 120 as shown in the figure. Find the magnetic induction at point P on the angle bisector is



A
7μ0M2πd3
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
7μ0M4πd3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
μ0M2πd3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
μ0M4πd3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 7μ0M2πd3
Since both M and 3M are making 120, therefore the resultant magnetic moment is,

Mres=M21+M22+2M1M2cosθ

=(M)2+(3M)2+2M(3M)cos120

=M2+9M2+6M2(12)

=7M2=7 M.

Now point P is on the axial line of the resultant magnetic moment.

By symmetry, it is clear that the resultant magnetic field is along the perpendicular bisector.

B=μ04π(2Md3)

B=μ04π(27 Md3)

B=7 μ0M2πd3

Hence, option (A) is the correct answer.


flag
Suggest Corrections
thumbs-up
4
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Dimensional Analysis
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon