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Q.

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 ( $μ_{0}$ = permeability for vacuum)

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a

$10^{- 17} μ_{0}$

b

$10^{- 7} μ_{0}$

c

$10^{- 6} μ_{0}$

d

$\frac{10^{- 7}}{μ_{0}}$

answer is B.

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Detailed Solution

$\begin{array}{l} i = \frac{q}{t} = 100 \times e \\ B_{centre} = \frac{μ_{0}}{4 π} . \frac{2 πi}{r} = \frac{μ_{0}}{4 π} = \frac{2 π \times 100 e}{r} \\ = \frac{μ_{0} \times 200 \times 1 .6 \times 10^{- 19}}{4 \times 0 .8} = 10^{- 17} μ_{0} \end{array}$

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