Consider a variation of Bohr’s 1913 model of the hydrogen atom, where the electr
ID: 1492675 • Letter: C
Question
Consider a variation of Bohr’s 1913 model of the hydrogen atom, where the electron is in a circular orbit of radius 4 × 1011 m, and its speed is 2.51625 × 106 m/s. What is the magnitude of the magnetic moment due to the electron’s motion? Answer in units of A m2 .
B) If the electron orbits counterclockwise in a horizontal circle as viewed from above upward, what is the direction of this magnetic moment vector?
1. radially directed; sometimes inward and sometimes outward
2. downward
3. radially outward
4. upward
5. radially inward
Explanation / Answer
the magnetic moment due to the electron’s motion is
u = I A = ( q/t) A = qv/ 2 pi r)( pir^2)
= 1/2 * q vr
= 1/2 ( 1.602 * 10^-19)(2.51625 × 10^6 m/s)(4 * 10^-11)
=8.062065 * 10^-24 A m^2
from the right hand rule , curl your fingers in the direction of the current as viewed from above. since current is the direction of motion of positive charges, the current is clockwise in a horizontal circle. your thumb should then point in the direction of the magnetic moment which is downward