CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A current 1A is flowing in the sides of equilateral triangle of side 4.5×102m .The magnetic field at centroid of the triangle is

23534_648a71f425fa4bd19d4fbfc5fea5b79e.png

A
4×105T
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
40T
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
0.4×105T
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
4×102T
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 4×105T
The triangle shown here can be considered to be having three finite wires of length a each. We need to find the magnetic field at a point on the perpendicular bisector of this wire.

The magnetic field due to a finite wire at a distance d from the wire, making angles θ1 and θ2 w.r.t the ends of the wire is given by

|B|=μ04πd(sinθ1+sinθ2)

This point is the same for all the three wires and at a distance of x from the wire. The direction of the magnetic field at that point is inwards due to all the wires. Thus, if B is the magnetic field due to one wire, then total magnetic field due to three wires will be 3B acting inwards


|B|=3×μ04πd(sinθ1+sinθ2)
The distance d can be expressed in terms of a by the formula da/2=tan30 . Substituting for d in the previous expression, we get,

|B|=3×μ04π(123)(sin60o+sin60o)
= 4×105T

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Parallel Currents
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon