Question

8. Electrons in the Hydrogen atom revolves 6.8×1015 times per second in a circular orbit of radius of 5.4×10-11m. Calculate the magnitude of magnetic field at the center of the orbit.​

Answer 1

Given : Electrons in the hydrogen atom revolves 6.8 × 10^15 times per second in a circular orbit of radius of 5.4 × 10¯¹¹ m.

To find : The magnitude of magnetic field at the centre of the orbit.

solution : magnetic field, B = $\frac{\mu_0i}{2R}$

i = q × f

= 1.6 × 10^-19 × 6.8 × 10^15

= 1.6 × 6.8 × 10¯⁴

= 1.088 × 10¯³ A

magnetic field , B = (4π × 10^-7 × 1.088 × 10¯³)/(2 × 5.4 × 10¯¹¹)

= (4 × 3.14 × 1.088 × 10^-10)/(1.08 × 10^-10)

= 12.65 T

The magnitude of magnetic field at the centre of the orbit is 12.65 T.

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