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Question
The given figure shows a long straight wire of a circular cross-section (radius a) carrying steady current I. The current I is uniformly distributed across this cross-section. Calculate the magnetic field in the region r < a and r > a.
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Solution
- Consider the case r > a. The Amperian loop, labelled 2, is a circle concentric with a cross-section. For this loop, L = 2πr
Using Ampere circuital Law, we can write,
B(2πr) = μ0I, `B = (mu_0I)/(2pir), B ∝ 1/r` (r > a) - Consider the case r < a. The Amperian loop is a circle labelled 1. For this loop, taking the radius of the circle to be r, L = 2πr
Now the current enclosed Ie is not I but is less than this value. Since the current distribution is uniform, the current enclosed is,
`I_e = I((pir^2)/(pia^2)) = (Ir^2)/a^2` Using Ampere’s law, B(2πr) = `mu_0 (Ir^2)/a^2`
B = `((mu_0I)/(2pia^2))r`
B ∝ r (r < a)
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