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प्रश्न
Using Ampere's circuital law, obtain an expression for the magnetic flux density 'B' at a point 'X' at a perpendicular distance 'r' from a long current-carrying conductor.
(Statement of the law is not required).
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उत्तर
Let AB, a long straight wire is carrying current i. The magnetic flux density `bar(B)` is to be required at point X at a distance r.
Let us draw a circle of radius r with centre O as shown in above figure. The point X lies on the circumference of the circle.
According to Ampere's circuital law,
`oint bar(B). bar(dl)= μ_0i. "where " bar(dl)` = infinitesimally small portion XY in the circle.
For complete circle, `bar(dl)= 2pir`
∴ `bar(B). 2pir = μ_0i`
⇒ `bar (B) = (mu_0)/(2pi).i/r NA^(-1)m^(- 1)`
⇒ `bar (B) = (mu_0)/(4pi).(2i)/r`
∴ `bar (B) = 10^-7 .(2i)/r NA^(-1)m^(- 1)`
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Substituting \[B=\frac{\mu_0 I}{2\pi r}\] into \[\oint\vec{B}\cdot d\vec{l}=B(2\pi r)\] for a straight wire yields:
