ISRO EC 2023 Q.65: solution and answer

ISRO 2023 · ECQ.65 · 1 mark · MCQ

Two infinitely long wire separated by a distance 5 m, carrying current I in opposite direction. If I = 10A, then the magnetic field intensity at point 'P' is

ISRO EC 2023 question 65: Two long parallel wires seen end-on, 5 m apart (a dimension line marked 5 m runs between them). — Figure from the ISRO EC 2023 paper. It is part of the question.

(A)
50/(8π) A/m
(B)
5/(8π) A/m
(C)
10/(8π) A/m
(D)
25/(8π) A/m

Answer and solution

ANSWER

$H = \frac{50}{8 π} A/m$ , option (a).

Official answer: ASource: ISRO's revised and final answer key of 16 February 2024

From the figure, P lies on the line joining the wires, $1 m$ from the right wire, so it is $5 - 1 = 4 m$ from the left wire.
For an infinitely long straight wire, $H = \frac{I}{2 π r}$ , with direction given by the right-hand rule.
Left wire (current into the page): the field circulates clockwise, so at a point to its right (P) it points downward.
Right wire (current out of the page): the field circulates counter-clockwise, so at a point to its left (P) it also points downward. The two fields point the same way and add.
$H = \frac{I}{2 π ( 1 )} + \frac{I}{2 π ( 4 )} = \frac{10}{2 π} (1 + \frac{1}{4})$
$H = \frac{10}{2 π} \cdot \frac{5}{4} = \frac{50}{8 π} A/m$
This is option (a). Subtracting the two fields, or using only one wire, gives the other options.

KEY IDEA

With currents in opposite directions, the fields of the two wires point the same way at any point between them, so their magnitudes $\frac{I}{2 π r}$ add.

WATCH OUT FOR

Subtracting the two fields (as for parallel currents), or taking the distance from the left wire as $1 m$ or $5 m$ instead of $4 m$ .

From the ISRO EC 2023 paper. Every question of this paper · The full paper (PDF, ISRO) · All 1 ISRO Magnetostatics question, with notes