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 Id placed in a magnetic field is given by
d F= Id l × B (16) The total force on a wire is the vector sum (integral) of the forces on all current elements.

Illustration–10
A wire is bent into a semicircular loop of radius R. It carries a current I, and its plane is perpendicular to a uniform magnetic field B, as shown in Fig (15 a).
Find the force on the loop. 
Solution
Consider a small element dl at an angle θ as shown in fig. (15 a).
The magnitude of force is dF = I dl B sin90° = IB dl = IBR dθ
The force can be resolved into components
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 :
here ∫d l 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 (16). 
Force of Interaction Between Parallel Wires
The interaction between two parallel wires can be summarised as follows :
 Like currents attract while unlike currents repel each other.
 The force of interaction per unit length is proportional to the product of the currents in each wire.
 The force is inversely proportional to the distance between them.
Consider two very long parallel wires carrying current I_{1}and I_{2}in the same direction as shown in the figure (17). The magnetic field produced by the current I_{1}at the position of I_{2}is given by
Illustration11
A rectangular loop of length l and width b carrying a current I_{2} is placed in the neighbourhood of a long strength wire carrying current I_{1} and shown in Fig. (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. 
Solution
(a) The force acting on each side of the loop are shown in 
(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
DRILL EXERCISE–3
 A wire carrying 15 A makes a 25º angle with a uniform magnetic field. The magnetic force per unit length of wire is 0.31 N/m. (a) What is the magnetic field strength? (b) What is the maximum force per unit length that could be achieved by reorienting the wire in this field?
 A piece of wire with mass per unit length 75 g/m runs horizontally at right angles to a horizontal magnetic field. A 6.2A current in the wire results in its being suspended against gravity. What is the magnetic field strength?
 Three parallel wires 4.6 m long each carry 20 A in the same direction. They’re spaced at the vertices of an equilateral triangle 3.5 cm on a side. Find the magnitude of the force on each wire.
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