JEE MAIN - Physics (2023 - 6th April Morning Shift - No. 11)
A long straight wire of circular cross-section (radius a) is carrying steady current I. The current I is uniformly distributed across this cross-section. The magnetic field is
uniform in the region $$r < a$$ and inversely proportional to distance $$r$$ from the axis, in the region $$r > a$$
zero in the region $$r < a$$ and inversely proportional to $$r$$ in the region $$r > a$$
directly proportional to $$r$$ in the region $$r < a$$ and inversely proportional to $$r$$ in the region $$r > a$$
inversely proportional to $$r$$ in the region $$r < a$$ and uniform throughout in the region $$r > a$$
Explanation
The magnetic field due to a current carrying wire can be calculated using Ampere's law. When the current is uniformly distributed across the cross-section of the wire, the situation will be different inside and outside the wire.
Inside the wire (r < a), the magnetic field is directly proportional to r (the distance from the center of the wire). This is because as you move away from the center of the wire, you enclose more current, so the magnetic field increases linearly with r.
Outside the wire (r > a), all the current in the wire is enclosed, so the magnetic field decreases with increasing r. This is a result of the magnetic field lines spreading out as they move away from the wire.
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