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प्रश्न
Two identical current carrying coaxial loops, carry current I in an opposite sense. A simple amperian loop passes through both of them once. Calling the loop as C ______.
- `oint B.dl = +- 2μ_0I`
- the value of `oint B.dl` is independent of sense of C.
- there may be a point on C where B and dl are perpendicular.
- B vanishes everywhere on C.
विकल्प
a and b
a and c
b and c
c and d
उत्तर
b and c
Explanation:
Ampere’s law gives another method to calculate the magnetic field due to a given current distribution.
Line integral of the magnetic field `vecB` around any closed curve is equal to μ0 times the net current i threading through the area enclosed by the curve, i.e.
`oint vecB * vec(dl) = mu_0 sumi = mu_0 (i_1 + i_3 - i_2)`
Total current crossing the above area is `(i_1 + i_3 - i_2)`. Any current outside the area is not included in net current. (Outward ⊙ → + ve, Inward ⊗ → – ve)
Applying the Ampere's circuital law, we have
`ointB* dl = i_0 (I - I) = 0` (because current is in opposite sense)
Also, there may be a point on C where B and dl are perpendicular and hence, `oint_c B*dl = 0`
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संबंधित प्रश्न
State Ampere’s circuital law.
Explain Ampere’s circuital law.
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(a) outside the cable
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(Statement of the law is not required).
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Using Ampere’s circuital law, obtain an expression for magnetic flux density ‘B’ at a point near an infinitely long and straight conductor, carrying a current I.
