First slide
Magnetic field due to a current carrying circular ring
Question

A particle carrying a charge equal to 100 times the charge on an electron is rotating per second in a circular path of radius 0.8 m. The value of the magnetic field produced at the center will be ( μ0 = permeability constant)

Easy
Solution

Given that
q=100e,r=0.8 and f=1/sec i=qt=qF=(100e×1)
Now B=μ0i2r=μ0(100e×1)2×0.8
=μ0×100×1.6×1019×11.6=1017μ0

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