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)
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a
10−7/μ0
b
10−17μ0
c
10−6/μ0
d
10−7μ0
answer is B.
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Detailed Solution
Given thatq=100e,r=0.8 and f=1/sec∴ i=qt=qF=(100e×1)Now B=μ0i2r=μ0(100e×1)2×0.8=μ0×100×1.6×10−19×11.6=10−17μ0