Question

P, Q and R long parallel straight wires in air,
carrying currents as shown. The direction of
resultant force on Ris
P Р
RI
2A 4A
6A
11)
(1) towards left
(2) towards right
(3) the same as that of current in Q
(4) perpendicular to plane of paper​

Answer 1

Answer:

(1) towards left is the correct answer.

Explanation:

Force between two wires carrying current in the same direction experiences attractive force toward each other.

This force has the magnitude = $\frac{\mu_o I_1 I_2}{2\pi d}$

Where, $I_1 , I_2$ are current in two wires respectively.

and, $d$ is the distance between the two parallel wires.

Now, from the figure we can see that all three of the wires carry current in the same direction, which means each wires is applying an attractive force on the other two wires.

For the wire R C carrying current 6A, wire Q applies an attractive force toward the left side, also wire P carrying 2A current also applies an attractive force towards the left side.

So, the net force on the wore R will be sum of these two attractive forces and net force on R will be towards the left.

#SPJ2

Answer 2

Answer:

Answer is Option (1) towards left

Explanation:

This force has the magnitude = $\frac{I_{1}I_{2}}{2\pi d}$

Where, $I_{1}I_{2}$ are current in two wires respectively.and, d  is the distance between the two parallel wires.

Now, from the figure we can see that all three of the wires carry current in the same direction, which means each wires is applying an attractive force on the other two wires.

For the wire R C carrying current $6$A, wire Q applies an attractive force toward the left side, also wire P carrying $2$A current also applies an attractive force towards the left side.

So, the net force on the wore R will be sum of these two attractive forces and net force on R will be towards the left.