Physics, 30.03.2020 20:25 Garciaapril1597

An electron and a proton move in circular orbits in a plane perpendicular to a uniform magnetic field B > . Find the ratio of the radii of their circular orbits when the electron and the proton have (a) the same momentum and (b) the same kinetic energy.

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Answer from: Esteffyy

$\frac{mv}{qB}$

a) ratio of radii for same momentum = 1

b) ratio of radii for same energy = 2.3 × $10^{-2}$

Explanation:

using formula for radius

r = $\frac{mv}{qB}$

where m is mass

v is velocity

q is charge

B is magnetic field

a) for same momentum radius will be inversely proportional to roduct of charge and magnetic and magnetic field

r ∝ $\frac{1}{qB}$

since for proton and electron s charge and magnetic field is same so

$\frac{r_{1} }{r_{2} } =\frac{1}{1}$

b) using relation b/w kinetic energy and momentum

p = $\sqrt{2 KEm}$ ............(1)

where p is momentum and

K.E is kinetic energy

we know $r =\frac{mv}{qB}$

mv =qBr

p = mv = qBr

from first equation

$qBr = \sqrt{2KEm}$

since qB is same for electron and proton and energy is also same(given)

so we get

$\frac{r_{e} }{ r_{p} }$ $= \sqrt{\frac{m_{e} }{m_{p } }$

$\frac{r_{e} }{r_{p} }$ $= \sqrt{\frac{9.1 x 10^{-31} }{1.67 x 10^{-27} } } = 2.3 x 10 ^{-2}$

Answer from: Quest

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Answer from: Quest

This is the answer: body 1 + engine 2

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