JEE Mains · Physics · STD 12 - Moving charges and magnetism
- JEE Mains 2026
- Hard
- MCQ
Two identical long current carrying wires are bent into the shapes shown in the following figures. If the magnitude of magnetic fields at the centres P and Q of a semicircular arc are \(B_1\) and \(B_2\) respectively, then the ratio \(\dfrac{B_1}{B_2}\) is _______.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(\dfrac{2 + \pi}{1 + \pi}\)
Step-by-step Solution
Detailed explanation
For the first figure, the magnetic field at the centre P is the vector sum of the magnetic fields due to two semi-infinite straight wires and one semicircular arc.
The magnetic field due to the top semi-infinite wire at P is \(B_{1a} = \dfrac{\mu_0 I}{4\pi r}\) directed into the plane.
The magnetic field due to the bottom semi-infinite wire at P is \(B_{1b} = \dfrac{\mu_0 I}{4\pi r}\) directed into the plane.
The magnetic field due to the semicircular arc at P is \(B_{1c} = \dfrac{\mu_0 I}{4r} = \dfrac{\pi \mu_0 I}{4\pi r}\) directed into the plane.
The total magnetic field at P is \(B_1 = B_{1a} + B_{1b} + B_{1c} = \dfrac{\mu_0 I}{4\pi r} (1 + 1 + \pi) = \dfrac{\mu_0 I}{4\pi r} (2 + \pi)\).
For the second figure, the magnetic field at the centre Q is the vector sum of the magnetic fields due to one horizontal semi-infinite wire, one semicircular arc, and one vertical semi-infinite wire.
The magnetic field due to the horizontal semi-infinite wire at Q is \(B_{2a} = \dfrac{\mu_0 I}{4\pi r}\) directed into the plane.
The magnetic field due to the semicircular arc at Q is \(B_{2c} = \dfrac{\mu_0 I}{4r} = \dfrac{\pi \mu_0 I}{4\pi r}\) directed into the plane.
Since the arc is semicircular, it ends at a point directly below Q. The vertical semi-infinite wire starts from this point and extends downwards, meaning its line of action passes through Q. Therefore, the magnetic field due to this vertical wire at Q is \(B_{2b} = 0\).
The total magnetic field at Q is \(B_2 = B_{2a} + B_{2b} + B_{2c} = \dfrac{\mu_0 I}{4\pi r} (1 + 0 + \pi) = \dfrac{\mu_0 I}{4\pi r} (1 + \pi)\).
The ratio of the magnetic fields is \(\dfrac{B_1}{B_2} = \dfrac{\dfrac{\mu_0 I}{4\pi r} (2 + \pi)}{\dfrac{\mu_0 I}{4\pi r} (1 + \pi)} = \dfrac{2 + \pi}{1 + \pi}\).
Answer: \(\dfrac{2 + \pi}{1 + \pi}\)
The magnetic field due to the top semi-infinite wire at P is \(B_{1a} = \dfrac{\mu_0 I}{4\pi r}\) directed into the plane.
The magnetic field due to the bottom semi-infinite wire at P is \(B_{1b} = \dfrac{\mu_0 I}{4\pi r}\) directed into the plane.
The magnetic field due to the semicircular arc at P is \(B_{1c} = \dfrac{\mu_0 I}{4r} = \dfrac{\pi \mu_0 I}{4\pi r}\) directed into the plane.
The total magnetic field at P is \(B_1 = B_{1a} + B_{1b} + B_{1c} = \dfrac{\mu_0 I}{4\pi r} (1 + 1 + \pi) = \dfrac{\mu_0 I}{4\pi r} (2 + \pi)\).
For the second figure, the magnetic field at the centre Q is the vector sum of the magnetic fields due to one horizontal semi-infinite wire, one semicircular arc, and one vertical semi-infinite wire.
The magnetic field due to the horizontal semi-infinite wire at Q is \(B_{2a} = \dfrac{\mu_0 I}{4\pi r}\) directed into the plane.
The magnetic field due to the semicircular arc at Q is \(B_{2c} = \dfrac{\mu_0 I}{4r} = \dfrac{\pi \mu_0 I}{4\pi r}\) directed into the plane.
Since the arc is semicircular, it ends at a point directly below Q. The vertical semi-infinite wire starts from this point and extends downwards, meaning its line of action passes through Q. Therefore, the magnetic field due to this vertical wire at Q is \(B_{2b} = 0\).
The total magnetic field at Q is \(B_2 = B_{2a} + B_{2b} + B_{2c} = \dfrac{\mu_0 I}{4\pi r} (1 + 0 + \pi) = \dfrac{\mu_0 I}{4\pi r} (1 + \pi)\).
The ratio of the magnetic fields is \(\dfrac{B_1}{B_2} = \dfrac{\dfrac{\mu_0 I}{4\pi r} (2 + \pi)}{\dfrac{\mu_0 I}{4\pi r} (1 + \pi)} = \dfrac{2 + \pi}{1 + \pi}\).
Answer: \(\dfrac{2 + \pi}{1 + \pi}\)
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