Force On A Current Carrying Conductor

Magnetics of Class 12

When a wire is placed in a magnetic field, it experiences no force. The thermal velocities of the free electrons are randomly oriented and so net force on them is zero. However, when a current flows, the electrons as a whole acquire a velocity in a definite direction and experience a magnetic force which is then transmitted to the wire.

The force experienced by an infinitesimal current element IdForce On A Current Carrying Conductor placed in a magnetic field Force On A Current Carrying Conductor is given by

dForce On A Current Carrying Conductor = IdForce On A Current Carrying Conductor × Force On A Current Carrying Conductor (3.16)

The total force on a wire is the vector sum (integral) of the forces on all current elements.

∫dForce On A Current Carrying Conductor = ∫(IdForce On A Current Carrying Conductor × Force On A Current Carrying Conductor)

Force On A Current Carrying Conductor

Example: 3.10

A wire is bent into a semi-circular loop of radius R. It carries a current I, and its plane is perpendicular to a uniform magnetic field B, as shown in
Fig (3.15 a).

Find the force on the loop.

Force On A Current Carrying Conductor

Solution

Consider a small element dl at an angle θ as shown in fig. (3.15 a).

The magnitude of force is dF = I dl B sin90° = IB dl = IBR  θ

The force can be resolved into components

dFx = dF cos θ = IBR cos θ dθ (i) dFy = dF sin θ = IBR sin θ dθ (ii) On integrating (i) and (ii) , we get

Fx = IBR Force On A Current Carrying Conductor

Fy = IBR Force On A Current Carrying Conductor

Force On A Current Carrying Conductor

Note that it is equal to the force acting on a straight wire of length 2R.

IMPORTANT

If the magnetic field is uniform and the current is constant, then both can be removed from the integral :

∫dForce On A Current Carrying Conductor = (I ∫dForce On A Current Carrying Conductor) × Force On A Current Carrying Conductor (3.17)

here ∫dForce On A Current Carrying Conductor indicates the vector sum of all the individual elements. This can be interpreted as follows :

The force on a curved wire joining point 1 and 2 is the same as that on a straight wire joining these points, as shown in fig (3.16).

Force On A Current Carrying Conductor

Force of Interaction Between Parallel Wires

The interaction between two parallel wires can be summarised as follows :

1. Like currents attract while unlike currents repel each other.

2. The force of interaction per unit length is proportional to the product of the currents in each wire.

3. The force is inversely proportional to the distance between them.

Consider two very long parallel wires carrying current I1 and I2 in the same direction as shown in the figure (3.17). The magnetic field produced by the current I1 at the position of I2 is given by

B21 = Force On A Current Carrying Conductor

Force On A Current Carrying Conductor

Note that the direction of Force On A Current Carrying Conductor21 is perpendicular to wire 2. So the magnitude of force on wire 2 is

Force On A Current Carrying Conductor21 = I2 (dForce On A Current Carrying Conductor × Force On A Current Carrying Conductor21) ⇒ F21 = I2 l B21 sin90°

or F21 = I2l Force On A Current Carrying Conductor

or F21 = Force On A Current Carrying Conductor

By applying the right hand rule one can note that the force is toward wire 1. The same analysis can be performed to find the force on the wire 1.

Thus, the force of interaction per unit length between two parallel wires is

Force On A Current Carrying Conductor (3.18)


Example: 3.11

A rectangular loop of length l and width b carrying a current I2 is placed in the neighbourhood of a long strength wire carrying current I1 and shown in
Fig. (3.18).

(a) Find the net force acting on the loop

(b) Find the work done to increase the spacing between the loop and the wire from a to 2a.

Force On A Current Carrying Conductor

Solution

(a) The force acting on each side of the loop are shown in
fig. (3.18 b) . Obviously, the magnitude of F2 is equal to that of F4. Thus, there is no net vertical force on the loop. The magnitude of F1 and F3 are :

F1 = Force On A Current Carrying Conductor

F2 = Force On A Current Carrying Conductor

The net force on the loop is horizontal and it is attractive.

Fnet = F1 − F2 = Force On A Current Carrying Conductor

Force On A Current Carrying Conductor

(b) If F is the instantaneous force acting on the loop when its separation form the wire is x, then the work down to increase the spacing from a to 2a is

W = Force On A Current Carrying Conductor

where F = Force On A Current Carrying Conductor

Thus, W = Force On A Current Carrying Conductor

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