Question

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.
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help please

Answer 1

Answer:

An electron moving around the nucleus is equivalent to a current,

Magnetic field at the centre,

B=μ0i2R=μ0qf2R

Substituting the values, we have

B=(4π×10−7)(1.6×10−19)(6.8×1015)2×5.1×10−11

=13.4T

b. The current carrying circular loop is equaivalent to a magnetic dipole with magnetic dipole moment,

M=NiA=(NqfπR2)

Substituting the values, we have

M=(1)(1.6×1019)(6.8×1015)(3.14)(5.1×10−11)2

=8.9×10−24A−m2

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