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Figure shows a rectangular 20-turn coil of wire, of dimensions 10cmby 5.0cm. It carries a current of 010Aand is hinged along one long side. It is mounted in the x-yplane, at angle θ=300to the direction of a uniform magnetic field of magnitude 0.50T. In unit-vector notation, what is the torque acting on the coil about the hinge line?

Short Answer

Expert verified

The torque acting on the coil about the hinge line is -4.3×10-3N - mj^.

Step by step solution

01

Write the given data

a) Number of turns of coil, N=20

b) The length of coil, l=10cm or0.10m

c) The breadth of coil, b=5cm or0.05m

d) The current through the coil, i=0.10A

e) The angle between the coil and direction of magnetic field is, θ'=30°

f) The magnetic field is, B=0.50T

02

Determine the formula for the torque

The area of the rectangle is as follows:

A=l×b …… (i)

Here, length of the rectangle, is the breadth of the rectangle.

The torque acting at a point inside a magnetic field is,

τ=NiABsinθ …… (ii)

Here, Nis the number of turns in the coil, iis the current of the wire, A is the area of the conductor, Bis the magnetic field, θis the angle made by the conductor with the magnetic field.

03

Determine the torque acting on the coil about the hinge line

The current passing through the segment AD which is along

the hinge line is in the +y direction, and the magnetic field is in the x-z plane. So, the torque acting on the coil will be in the –y direction.

Here,θis the angle between magnetic dipole moment and the magnetic field.

So, θ=90-θ'=60°

Also, the area of rectangular coil can be given using the given data in equation (i) as follows:

A=0.10m×0.05m=0.005m2

Substituting all the known values in the equation (ii), we get the torque value about the hinge line as follows:

τ=20×0.10A×0.005m2×0.50T×sin600=4.3×10-3N - m

As, this torque is in the –y direction, we can write it in unit vector notation as,

τ→=-4.3×10-3N - mj^

Hence, the required value of the torque is -4.3×10-3N - mj^.

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