Magnetic Effects & Magnetism

JEE Main level, single correct, medium

272 of 276

Question 272 of 276

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 B1B_1 and B2B_2 respectively, then the ratio B1B2\dfrac{B_1}{B_2} is _______.

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Answer: (A) 2+π1+π\dfrac{2 + \pi}{1 + \pi}

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 B1a=μ0I4πrB_{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 B1b=μ0I4πrB_{1b} = \dfrac{\mu_0 I}{4\pi r} directed into the plane.

The magnetic field due to the semicircular arc at P is B1c=μ0I4r=πμ0I4πrB_{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 B1=B1a+B1b+B1c=μ0I4πr(1+1+π)=μ0I4πr(2+π)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 B2a=μ0I4πrB_{2a} = \dfrac{\mu_0 I}{4\pi r} directed into the plane.

The magnetic field due to the semicircular arc at Q is B2c=μ0I4r=πμ0I4πrB_{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 B2b=0B_{2b} = 0.

The total magnetic field at Q is B2=B2a+B2b+B2c=μ0I4πr(1+0+π)=μ0I4πr(1+π)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 B1B2=μ0I4πr(2+π)μ0I4πr(1+π)=2+π1+π\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: 2+π1+π\dfrac{2 + \pi}{1 + \pi}

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