Question

As shown in the figure 20A, 40A and 60A currents are passing through very long straight wires P, Q and R respectively in the directions shown by the arrows. In this condition the direction of the resultant force on wire Q is(A) towards left of wire Q(B) towards right of wire Q (C) normal to the plane of paper(D) in the direction of current passing through Q.

Answer 1

answer : (A)  toward left of wire Q

explanation:- we know current carrying two parallel wires attract each other when current through each wire is in same direction. and repel each other when current through wire is in opposite direction.

in case of  wire P and wire Q ,

direction of current is in same direction so the force will be attractive in nature and we know force is directly proportional to product of currents .

so, F = Ki₁i₂ = K.20.40 = 800K N (attractive, ∴ direction of force towards left direction)

similarly, in case of wire Q and wire R,

direction of current is in opposite direction so, the force act between them will be repulsive.

F' = Ki₂i₃ = K.40.60 = 2400K N ( repulsive force, ∴ direction of force towards left direction)

so, Fnet = F + F' = 2400K + 800K = 3200K N (left direction)

hence, force on Q towards left of wire Q.