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

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10−7μ0

10−17 μ0

10−6 μ0

10−7 μ0

answer is B.

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Detailed Solution

i=qt = 100 × eBcentre = μ04π . 2πir = μ04π = 2π × 100er= μ0 × 200 × 1.6 × 10−194 × 0.8 = 10−17 μ0

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