JEE Main 2026 — Jan 28, Shift 1 — Physics solution

JEE Main2026Jan 28, Shift 1Physics

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 ______ .

A
$6 \times 1 0^{- 7}$ N towards $P$
B
$6 \times 1 0^{- 6}$ N towards $R$
C
$6 \times 1 0^{- 7}$ N towards $R$
D
$6 \times 1 0^{- 6}$ N towards $P$

Solution

Force per unit length between two parallel currents: $\frac{F}{ℓ} = \frac{μ _{0} I _{1} I _{2}}{2 π d}$ . Antiparallel currents repel; parallel currents attract.

P (3 A up) and Q (1 A down) are antiparallel $\Rightarrow$ Q is repelled by P, pushed toward R.

R (2 A down) and Q (1 A down) are parallel $\Rightarrow$ Q is attracted to R, also pulled toward R.

Both forces on Q point toward R. Magnitudes (per unit length):

From P: $\frac{( 2 \times 1 0 ^{- 7} ) ( 3 ) ( 1 )}{0.03} = 2 \times 1 0^{- 5}$ N/m.

From R: $\frac{( 2 \times 1 0 ^{- 7} ) ( 1 ) ( 2 )}{0.02} = 2 \times 1 0^{- 5}$ N/m.

Total per metre: $4 \times 1 0^{- 5}$ N/m. Over 15 cm = 0.15 m: $F = 4 \times 1 0^{- 5} \times 0.15 = 6 \times 1 0^{- 6}$ N, directed toward R.

Option (2).

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