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goblinko [34]
3 years ago
7

To get an idea of the size of magnetic fields at the atomic level, consider the magnitude of the magnetic field due to the elect

ron in the simple Bohr model of the hydrogen atom. In the ground state the Bohr model predicts that the electron speed would be 2.2 × 106 m/s, and the distance from the proton would be 0.5 × 10−10 m. What is at the location of the proton?
Physics
1 answer:
artcher [175]3 years ago
3 0

Answer:

The magnetic field is 14.08 T.

Explanation:

Given that,

Speed of electron v=2.2\times10^{6}\ m/s

Distance d =0.5\times10^{-10}\ m

Suppose we need to find the magnetic field at  the location of the proton

We need to calculate the magnetic field

Using formula of magnetic field

B = \dfrac{\mu_{0}I}{2r}....(I)

Using formula of current

I=\dfrac{q}{T}

Using formula of time

T= \dfrac{2\pi r}{v}

Put the value of current and time in equation (I)

B=\dfrac{\mu \times q\times v}{2\pi r\times2r}

Put the value into the formula

B=\dfrac{4\pi\times10^{-7}\times1.6\times10^{-19}\times2.2\times10^{6}}{4\pi\times(0.5\times10^{-10})^2}

B=14.08\ T

Hence, The magnetic field is 14.08 T.

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The length of a displaced pendulum body which passes its lowest point twice every second
Troyanec [42]

Answer:

0.248 m

Explanation:

The period of a simple pendulum is given by

T=2\pi \sqrt{\frac{L}{g}}

where

L is the length of the pendulum

g is the acceleration of gravity

The pendulum in the problem passes its lowest point twice every second: this means: this means that it makes one complete oscillation in one second, so its period is 1 second:

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And using

g=9.8 m/s^2

We can rearrange the equation above to find L, the length of the pendulum:

L=\frac{T^2 g}{4\pi^2}=\frac{1^2(9.8)}{4\pi^2}=0.248 m

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