Question

A circular loop of wire with a radius of 15 cm lies in the yz plane and carries a current of 1.9 A, clockwise when viewed from a point on the positive x axis. The magnitude and direction of its magnetic dipole moment are: Select one: a. 1343.03 A m^2, -x direction b. 0.13 A m^2, +x direction c. 1343.03 A m^2, +x direction d. 0.13 A m^2, -x direction e. 0.13 A m^2, +z direction

Answer 1

Magnetic dipole moment M = IA

Where I- current

A- area of the loop

Here our loop is in the shape of circle. So , area of circle = Πr^2

Given that r = 15 cm = 0.15 m

Current I= 1.9 A ( clockwise )

Hence M = IA= 1.9 x Π(0.15)^2 = 0.1343 A m^2

Or on rounding off

M= 0.13 A m^2

Coming to its direction it is found out using Right hand screw principle. If if we curl our right hand fingers except thumb in the direction of the current flowing that is clockwise direction then the thumb show the direction of the magnetic moment.

It is shown in the below figure,

न T 2. -X C.

Here Loop is in the plane of yz the current is clockwise hence the direction of magnetic moment is along -ve x- axis.

Hence option d is correct.

I.e. 0.13 A m^2 - x direction.

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