GATE 2026 IN Q.38: solution and answer

GATE 2026 · INQ.38 · 2 marks · MCQ

A straight thin wire carrying a current of $I = 1$ A in the direction $\overset{n}{^} = \frac{1}{6} \hat{i} + \frac{2}{6} \hat{j} + \frac{1}{6} \hat{k}$ is placed inside a magnetic field of $B = (2 \hat{i} + 2 \hat{k})$ T. The force per unit length on the wire is ______ N/m.

(A)
$26 \hat{i} - 26 \hat{k}$
(B)
$\frac{3}{2} \hat{i} - \frac{3}{4} \hat{j} + \frac{3}{4} \hat{k}$
(C)
$\frac{4}{6} \hat{i} - \frac{4}{6} \hat{k}$
(D)
$\frac{3}{2} \hat{i} + \frac{3}{4} \hat{j} + \frac{3}{4} \hat{k}$

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Answer and solution

ANSWER

The force per unit length is $\frac{F}{L} = I \overset{n}{^} \times B = \frac{4}{6} \hat{i} - \frac{4}{6} \hat{k} N/m$ , which is option (C).

Official answer: CSource: Official answer key, GATE 2026 IN (IIT Guwahati)

The force on a straight wire of length $L$ carrying current $I$ in a uniform field is $F = I L \times B$ with $L = L \overset{n}{^}$ . So the force per unit length is $\frac{F}{L} = I \overset{n}{^} \times B$ , with $I = 1 A$ .
In components, $\overset{n}{^} = \frac{1}{6} (1, 2, 1)$ and $B = (2, 0, 2) T$ . Check: $∣ \overset{n}{^} ∣ = (1 + 4 + 1) /6 = 1$ , so $\overset{n}{^}$ is already a unit vector and needs no further normalising.
Write the cross product as a determinant: $\overset{n}{^} \times B = \frac{1}{6} \hat{i} 12 \hat{j} 20 \hat{k} 12$
Expand it: $\hat{i} (2 \cdot 2 - 1 \cdot 0) - \hat{j} (1 \cdot 2 - 1 \cdot 2) + \hat{k} (1 \cdot 0 - 2 \cdot 2) = 4 \hat{i} + 0 \hat{j} - 4 \hat{k}$ .
So, with $I = 1 A$ : $\frac{F}{L} = \frac{4}{6} \hat{i} - \frac{4}{6} \hat{k} N/m$
Check the direction. The force must be perpendicular to both the wire and the field: $(4, 0, - 4) \cdot (1, 2, 1) = 4 - 4 = 0$ and $(4, 0, - 4) \cdot (2, 0, 2) = 8 - 8 = 0$ . Both dot products are zero.
Check the magnitude with $∣ F ∣/ L = I ∣ B ∣ sin θ$ . Here $∣ B ∣ = 22$ and $cos θ = \frac{n ^ \cdot B}{∣ B ∣} = \frac{4/ 6}{2 2} = \frac{1}{3}$ , so $sin θ = 2/3$ . That gives $∣ F ∣/ L = 22 \cdot 2/3 = \frac{4}{3} N/m$ , the same as $(4/ 6)^{2} + (4/ 6)^{2} = 16/3$ .
Rule out the other options. The force has no $\hat{j}$ component, which excludes (B) and (D); neither of them is perpendicular to $B$ anyway. (A) points the same way as (C) but is 3 times larger, since $26 = 12/ 6$ . Its magnitude, $43 \approx 6.93 N/m$ , is more than the largest possible value $I ∣ B ∣ = 22 \approx 2.83 N/m$ . Only (C) is left.

KEY IDEA

For a straight wire in a uniform field, $F / L = I \overset{n}{^} \times B$ . The force is perpendicular to both the wire and the field, and its size can never be more than $I ∣ B ∣$ . Those two facts rule out wrong options before you finish the cross product.

WATCH OUT FOR

Common mistakes are reversing the order to $B \times \overset{n}{^}$ , which flips the sign to $- \frac{4}{6} \hat{i} + \frac{4}{6} \hat{k}$ , and dropping or mis-handling the $1/ 6$ factor in $\overset{n}{^}$ , which gives a vector in the right direction with the wrong size, like option (A). A quick check: the force must be perpendicular to $B$ , and its magnitude cannot be more than $I ∣ B ∣ = 22 N/m$ .

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GATE 2026 IN Q.38: solution and answer

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