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Is displacement current produce magnetic field? Why

Asked by ashutosharnold1998 | 12 May, 2019, 01:32: PM

Expert Answer

displacement current i_d = ε₀ dφ/dt =

begin mathsize 12px style epsilon subscript 0 fraction numerator d over denominator d t end fraction contour integral subscript S E. d a space space equals space epsilon subscript 0 space A space fraction numerator d E over denominator d t end fraction end style

..............(1)

where ε₀ is permitivity of free space, φ is electric field flux , E is electric field and A is closed surface considered in electric field .

If moving charges are causing conventional current, displacement current is caused by change of electric field.

Magnetic field of displacement current is obtained by writing Amphere's law only for displacement current,

begin mathsize 12px style contour integral subscript C B. d s space equals space mu subscript 0 space i subscript d space space space o r space space space B space equals space space mu subscript 0 space epsilon subscript 0 space A over l space fraction numerator d E over denominator d t end fraction space equals space 4 straight pi cross times 10 to the power of negative 7 end exponent cross times 8.85 cross times 10 to the power of negative 12 end exponent straight A over straight l space dE over dt space almost equal to space 10 to the power of negative 17 end exponent straight A over straight l space dE over dt end style

.........................(2)

In eqn.(2) l is loop length considered to get magnetic field at the points of loop.

From eqn.(2), we see that to get appreciable magnetic field, either A/l or dE/dt should be very large.

Answered by Thiyagarajan K | 12 May, 2019, 03:44: PM

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