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A long straight wire of radius 'a' carries a steady current 'I'. The current is uniformly distributed across its area of cross-section. The ratio of the magnitude of magnetic field

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A long straight wire of radius 'a' carries a steady current 'I'. The current is uniformly distributed across its area of cross-section. The ratio of the magnitude of magnetic field `vecB_1` at `a/2` and `vecB_2` at distance 2a is ______.

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  • `1/2`

  • 1

  • 2

  • 4

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Solution

A long straight wire of radius 'a' carries a steady current 'I'. The current is uniformly distributed across its area of cross-section. The ratio of the magnitude of magnetic field `vecB_1` at `a/2` and `vecB_2` at distance 2a is 1.

Explanation:

Consider two amperian loops of radius `a/2` and 2a. Applying Ampere's circuital law for these loops.

We get `ointB.dL = μ_0I_{"enclosed"}`

For the smaller loop,

`B_1 xx 2pi a/2 = mu_0 xx pia^2 1 xx (a/2)^2`

`mu_0I xx 1/4 = (mu_0I)/4`

`B_1 = (mu_0I)/(4pia)`

`B_2 xx 2pi(2a) = mu_0I`

`B_2 = (mu_0I)/(4pia)`

∴ B1/B2 = `(mu_0I)/(4pia) xx (4pia)/(mu_0I) = 1`

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