An electron is revolving around a proton in a circular orbit of diameter 1A∘. If it produces a magnetic field of 14 Wb/m2 at the proton, then its angular velocity will be about

 In a circular orbit with a diameter of 1A, an electron revolves around a proton. If we are asked to calculate the angular velocity of a proton with a magnetic field of 14 Wb/m2, we must first grasp the link between current, electron, and time.

Current passing through the wire is equal to the number of electrons passing through the wire in time t, this can be shown mathematically as,

I = q/t …………..(i)

Where i is current, q is electric charge and t is time.

Now, we know that the relationship between time and angular velocity can be given by,

t=2π/w ………..(ii)

Where t is time and w is angular velocity.

Now, in question it is given that, the magnetic field produced due to the revolution of electron is 14 Wb/m2, so, we can say that the electron forms a circular loop around proton while revolving and we know that magnetic field produced due to the circular loop can be given by the formula,

B = μ0i/2R ………………….(iii)

Where i is current, B is the magnetic field and R is the radius, μ0=4π×10−7 .

Now, the value of the magnetic field is 14 Wb/m2 and diameter is 1A∘, substituting these values in the expression we will get,

B = μ0q/2Rt from expression (i)

⇒B = μ0qw/2R2π from expression (ii)

Now, substituting all the given values we will get,

⇒14 = 4π×10−7qw/2D/22π


As, R=D2 and 1A=10−10 , now, q = e− =1.6×10−19 , so, now the expression will be,

⇒14=10−7× 1.6×10−19×w/10−10


⇒ w = 8.75×1016 rad/s

Hence, angular velocity is 8.75×1016 rad/s

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