A long straight wire with a circular cross-section having radius R, is carrying a steady current I. The current I is uniformly distributed across this cross-section. Then the variation of magnetic field due to current I with distance r (r < R) from its centre
- \(B∝r^2\)
- \(B∝r\)
- \(B∝\frac {1}{r^2}\)
- \(B∝\frac {1}{r}\)
The Correct Option is B
Solution and Explanation
We know that,
\(∫B.dl = μ_0l_{in}\)
\(B \times 2\pi r = \frac {μ_0l}{\pi R^2} \times \pi r^2\)
\(⇒ B ∝ r\)
So, the correct option is (B): \(B ∝ r\)
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Concepts Used:
Amperes circuital law
According to Ampere’s law, magnetic fields are related to the electric current that is produced in them. This law specifies that the magnetic field is associated with a given current or vice-versa, provided that the electric field doesn’t change with time.
Ampere’s circuital law can be written as the line integral of the magnetic field surrounding the closed loop which is equal to the number of times the algebraic sum of currents passing through the loop.
According to the second equation, if the magnetic field is integrated along the blue path, then it is equal to the current enclosed, I.
The magnetic field doesn’t vary at a distance r because of symmetry. The path length (in blue) in figure 1 has to be equal to the circumference of a circle,2πr.
