Q132 / 150 · MHT CET 2026 (11 April Shift 2) · Physics
Two parallel wires of equal lengths are separated by a distance of $3$ m from each other. The currents flowing through first and second wire is $3$ A and $4.5$ A respectively in opposite directions. The resultant magnetic field at mid point of both the wires is ($\mu_0$=permeability of free space)
- A$\dfrac{\mu_0}{2\pi}$
- Correct answerB$\dfrac{5\mu_0}{2\pi}$
- C$\dfrac{7\mu_0}{2\pi}$
- D$\dfrac{9\mu_0}{2\pi}$
Solution
Distance between the wires, $d = 3$ m.
Distance of the midpoint from each wire, $r = \dfrac{d}{2} = 1.5$ m.
Current in the first wire, $I_1 = 3$ A.
Current in the second wire, $I_2 = 4.5$ A.
Since the currents are flowing in opposite directions, the magnetic fields produced by both wires at the midpoint will be in the same direction.
The resultant magnetic field at the midpoint is $B = B_1 + B_2$.
$B = \dfrac{\mu_0 I_1}{2\pi r} + \dfrac{\mu_0 I_2}{2\pi r}$
$B = \dfrac{\mu_0}{2\pi r} (I_1 + I_2)$
Substituting the given values:
$B = \dfrac{\mu_0}{2\pi (1.5)} (3 + 4.5)$
$B = \dfrac{\mu_0}{2\pi (1.5)} (7.5)$
$B = \dfrac{5\mu_0}{2\pi}$
Answer: $\dfrac{5\mu_0}{2\pi}$
Distance of the midpoint from each wire, $r = \dfrac{d}{2} = 1.5$ m.
Current in the first wire, $I_1 = 3$ A.
Current in the second wire, $I_2 = 4.5$ A.
Since the currents are flowing in opposite directions, the magnetic fields produced by both wires at the midpoint will be in the same direction.
The resultant magnetic field at the midpoint is $B = B_1 + B_2$.
$B = \dfrac{\mu_0 I_1}{2\pi r} + \dfrac{\mu_0 I_2}{2\pi r}$
$B = \dfrac{\mu_0}{2\pi r} (I_1 + I_2)$
Substituting the given values:
$B = \dfrac{\mu_0}{2\pi (1.5)} (3 + 4.5)$
$B = \dfrac{\mu_0}{2\pi (1.5)} (7.5)$
$B = \dfrac{5\mu_0}{2\pi}$
Answer: $\dfrac{5\mu_0}{2\pi}$
Hence, the correct answer is option: 'B'.