## Wednesday, June 27, 2007

### Net Magnetic Force and Torque on a Current Loop- Two Multiple Choice Questions

If a plane current carrying coil is placed in a uniform magnetic field, generally there will be a torque on the coil, but the net force on the coil will be zero. But, if the current loop is placed in a non-uniform magnetic field, there will be a torque and a net force on the coil. The torque on the loop will be zero only if the plane of the loop is perpendicular to the direction of the magnetic field B. This follows from the expression for torque(τ) given by

τ = n IABsinθ, where ‘n’ is the number of turns, ‘A’ is the area of the loop, ‘I’ is the current and ‘θ’ is the angle between the magnetic field vector B and the area vector A. Note that the area vector is directed normal to the plane of the area. So, if the plane of the current loop is perpendicular to the direction of the magnetic field, the torque is zero. The torque is maximum (equal to nIAB) when the plane of the loop is parallel to the field B.

Now, consider the following MCQ:

A thin copper wire of length ‘L’ is bent to form a single turn plane circular loop and is suspended in a uniform magnetic field ‘B’ which is directed parallel to the plane of the loop. The torque acting on the loop when a current ‘I’ passes in it is 0.1 N. If the same wire were bent to form a plane circular loop of 10 turns, the torque would be

(a) 0.01 N (b) 0.1 N (c) 1 N (d) 0.001 N (e) 10 N

The torque is maximum (equal to nIAB) since the plane of the coil is parallel to the magnetic field. In the first case, since the number of turns (n) is one,

torque, τ = IAB = I×π(L/2π)2×B since the radius of the single turn coil is L/2π.

Therefore, τ = IL2B/4π

When the coil is of ‘n’ turns (with the wire of the same length L),

torque, τ’ = nIA’B = nI×π(L/2πn)2×B = IL2B/4πn.

The torque thus reduces to (1/n)th of the torque on the single turn coil. Since n = 10, the correct option is 0.01 N.

Let us now consider the following question which is popular among question setters: A very long straight wire carrying a current ‘I’ is arranged to be coplanar with a rectangular loop of sides ‘a’ and ‘b’ carrying a current ‘i’ as shown. If the nearer parallel side of the loop is at distance ‘r’ from the straight wire, the magnitude of the net magnetic force on the loop is

(a) (μ0I iab)/[2πr(r+b)]

(b) (μ0I iab)/[2πr(r+b/2)]

(c) (μ0I iab)/2πr

(d) (μ0I iab)/[2π(r+b)]

(e) (μ0I iab)/[πr(r+b)]

The magnetic forces on the top and bottom sides of the loop are equal and opposite and hence they will get canceled. The force on the left side of the loop is attractive (towards the straight wire) while the force on the right side of the loop is repulsive (away from the straight wire). But they cannot get canceled since the attractive force is greater because of the proximity of the left side to the straight wire.

The magnetic force on the left side = ia(μ0I/2πr) since (μ0I/2πr) is the magnetic flux density produced by the straight wire at the distance ‘r’.

Similarly, magnetic force on the right side = ia[μ0I/2π(r+b)]

Therefore, the net magnetic force on the loop is

ia(μ0I/2π)[1/r – 1/(r+b)] = 0Iiab)/[2πr(r+b)]

You will find many useful questions (with solution) on magnetic force at apphysicsresources