JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Magnetic Effects of Current

Three long straight wires carrying current are arranged mutually parallel as shown in the figure. The force experienced by 15 cm length of wire

Q

is ______.
(

μ_{0} = 4 π \times 1 0^{- 7} T . m / A

)

Answer: (B)

Solution

The force per unit length between two parallel current-carrying wires is

\frac{F}{ℓ} = \frac{μ _{0} I _{1} I _{2}}{2 π d}

, attractive for currents in the same direction and repulsive for opposite directions.

Wire

Q

carries

1 A

downward. Wire

P

(

3 A

upward,

3 cm

to the left) is antiparallel to

Q

, so it repels

Q

and pushes it to the right, towards

R

. Wire

R

(

2 A

downward,

2 cm

to the right) is parallel to

Q

, so it attracts

Q

, again pulling it towards

R

. The two forces act in the same direction and add.

For the

ℓ = 15 cm

length of

Q

F = \frac{μ _{0} ℓ I _{Q}}{2 π} (\frac{I _{P}}{d _{P}} + \frac{I _{R}}{d _{R}}) = 2 \times 1 0^{- 7} \times 0.15 \times 1 \times (\frac{3}{0.03} + \frac{2}{0.02})

F = 2 \times 1 0^{- 7} \times 0.15 \times (100 + 100) = 6 \times 1 0^{- 6} N, directed towards R .

JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Single correct option