Question

In niels bohr's 1913 model of the hydrogen atom an electron circles that proton at a distance of 5.29e-11 m with a speed of 2.19e6 m/s. compute the magnitude of themagnetic field that this motion produces at the location of the proton

Answer 1

Concepts and reason

The concepts required to solve this problem are Bohr’s model of the hydrogen atom and the magnetic field.

First, calculate the current by using the values of the radius of the circular path of the electron, speed of the electron, and the charge of the electron. Finally, calculate the value of the magnetic field by using the values of the current and the radius of the circular path of the electron.

Fundamentals

The expression for the current is,

Here, is the charge of the electron and is the period of the circular motion of the electron.

The expression for the period of the circular motion of the electron is,

Here, is the radius of the circular path of the electron and v is the speed of the electron.

The expression for the magnetic field inside the circular path is,

Here,is the permeability of the free space.

The expression for the current is,

Here, is the charge of the electron and is the period of the circular motion of the electron.

The expression for the period of the circular motion of the electron is,

Here, is the radius of the circular path of the electron and v is the speed of the electron.

Substitute forin equation.

Substitute for , for, and for.

The expression for the magnetic field inside the circular path is,

Here,is the permeability of the free space.

Substitute for, for , and for .

Ans:

The required magnetic field is.

Answer 2

The Proton is stationary so therefore no magnetic force is beingexerted by the proton on the electron. There is an electric fieldpresent. This probleminvolves the use of the Biot-Savartlaw(Uo=permeability of space).

B= (Uo*q*v/4*π*r^2).

The only variation made to the actual formula was the introductionof q &v instead of I.
I=n*A*q*v
Since there is only 1 carrier,(electron) the current equationreduces to q*v (v=drift speed). (n*A represent charge density in asection)

Here are the calculations:

B= ((Uo*1.6e-19*2.19e6)/(4*π*(5.29e-11^2)))= 12.55 T

Answer 3

Magnetic field in Circular motion of the eelctron B = mv/ R qwhere m = mass of electron = 9.1 * 10 ^-31kgv = speed = 2.19 * 10 ⁶ m / sR = distance = 5.29 * 10 ^-11 mq = charge = 1.6 * 10 ^-19 Csubstitue values weget B = 2.35 * 10 ⁵T

Answer 4

The answer above of 2.35*10^5 T is what I got, but unfortunately,the answer in the back of the book is 12.5 T

This question is in the section titled "The Biot-Savart Law".

Any ideas on how they got the answer of 12.5 T?

Answer 5

The electron circles the proton at a distance of r = 5.29 *10^-11 mThe speed of the electron is v = 2.19 * 10⁶m/sThe centripetal force on the electron isF = (mv²/r) ----------(1)The magnetic force on the electron isF_B= q * v * B ----------(2)Fromequations (1) and (2),we get(mv²/r) = (q * v * B)or B = (m * v/q * r)Here,m = 9.1 * 10^-31 kg and q = 1.6 *10^-19 Cor B = [(9.1 * 10^-31 * 2.19 * 10⁶/1.6 *10^-19* 5.29 * 10^-11 )or B = 2.35 * 10⁵ TThe magnitude of the magnetic field that this motion producesat the location of the proton is B = 2.35 * 10⁵ T.