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

# A coil in the shape of equilateral tringale of side 0.02 m is suspended from the vertex such that it is hanging in a vertical place between the pole-pieces of a permanent magnet producing a horizontal magnetic field of 5×10−2 T. When a current of 0.1 A passed through it and the magnetic field is parallel to its plane then couple acting on the coil is :

A
8.65×107 Nm
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
6.65×107 Nm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
3.35×107 Nm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
3.91×107 Nm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is A 8.65×10−7 N−mThe torque acting on a coil is given by,τ = N B I A sinθHere, A = 12×base× height=12×0.02×√0.022−0.012=1.732×10−4m2θ = Angle between direction of the magnetic field and normal to the plane of the coil. Also, as the magnetic field is parallel to the plane of the coil, so θ = 90thus, τ = 1×5×10−2×0.1×1.732×10−4×1 = 8.6×10−7 Nm

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Algebraic Identities
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program