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Q.

An electron revolves in a circle of radius $0 . 4^{0} A$ with a speed of $10^{5} m / s$ . The magnitude of the magnetic field, produced at the center of the circular path due to the motion of the electron in $W b / m^{2}$ is

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a

1 $W b / m^{2}$

b

2 $W b / m^{2}$

c

4 $W b / m^{2}$

d

3 $W b / m^{2}$

answer is A.

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Detailed Solution

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$M a g n e t i c f i e l d a t t h e c e n t e r o f c o i l, B = \frac{μ_{o} i}{2 r} \Rightarrow B = \frac{μ_{o} \frac{q}{T}}{2 r} \Rightarrow B = \frac{μ_{o} \frac{q}{\frac{2 π r}{V}}}{2 r} = \frac{μ_{o}}{4 π} \frac{q V}{r^{2}} = 10^{- 7} \frac{1.6 x 10^{- 19} x 10^{5}}{{(0.4 x 10^{- 10})}^{2}} = 1 W b / m^{2}$

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