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Q.

A particle carrying a charge equal to 100 times the charge on an electron, is rotating per sec in a circular path of radius 0.8 m. The value of the magnetic field produced at the centre will be :

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a

$10^{- 6} μ_{0}$

b

$10^{- 17} μ_{0}$

c

$10^{- 7} / μ_{0}$

d

$10^{- 7} μ_{0}$

answer is B.

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Detailed Solution

current due to the charge= $1.6 \times 10^{- 17}$ A

thus magnetic field at the center = $\frac{μ_{o} (1.6 \times 10^{- 17})}{2 \times 0.8}$ = $μ_{0} 10^{- 17}$

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A particle carrying a charge equal to 100 times the charge on an electron, is rotating per sec in a circular path of radius 0.8 m. The value of the magnetic field produced at the centre will be :