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 will be:

Option 1 -

B ∝ r²

Option 2 -

B ∝ r

Option 3 -

B ∝ $\frac{1}{r^{2}}$

Option 4 -

B ∝ $\frac{1}{r}$

4 Views | Posted 3 months ago

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1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

3 months ago

Correct Option - 2

Detailed Solution:

for $π R^{2} A r e a \to' l'$ $^{}$ current

1 unit Area $\frac{l}{π R^{2}}$ $\frac{}{^{}}$

$π r^{2} \to \frac{l}{π R^{2}} \times π r^{2}$

$i = \frac{l r^{2}}{R^{2}}$

Now, consider Amperian loop of radius small ‘r’ ln Amperian loop magnetic field will be tangential to the amperian loop.

$? \vec{B} . d \vec{i} = μ_{0} l_{e n c l o s e d}$ (Ampere circuital law)

$B = \frac{μ_{0}}{2 π} \frac{l}{R^{2}} r$

$B \propto r$

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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 will be:

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