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Expert-verified Found in: Page 91 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # A magnetic field forces an electron to move in a circle with radial acceleration ${\mathbf{3}}{\mathbf{.}}{\mathbf{0}}{\mathbf{×}}{{\mathbf{10}}}^{{\mathbf{14}}}{\mathbf{m}}{\mathbf{/}}{{\mathbf{s}}}^{{\mathbf{2}}}$ in a particular magnetic field. (a) What is the speed of the electron if the radius of its circular path is ${\mathbf{15}}{\mathbf{cm}}$? (b)What is the period of the motion?

a) Speed of the electron is $6.7×{10}^{6}m/s$.

b) Period of motion is$1.4×{10}^{-7}s$.

See the step by step solution

## Step 1: Given

1) Radial acceleration, $a=3×{10}^{14}m/{s}^{2}$

2) Radius of the circular path,

$r=15\mathrm{cm}\phantom{\rule{0ex}{0ex}}=0.15\mathrm{m}$

## Step 2: Understanding the concept of centripetal acceleration

The centripetal acceleration of the object depends on the velocity and radius of the orbit. Using the equation of centripetal acceleration, we can calculate the speed of an electron moving in a circular direction. Using the value of the speed of an electron and the equation of period, we can find the period of motion of an electron.

Formula:

$a=\frac{{V}^{2}}{r}$ …(i)

$T=\frac{2\pi r}{V}$ …(ii)

## Step 3: (a) Calculate the speed of the electron if the radius of its circular path is 15cm

From equation (i), we can write,

$a=\frac{{V}^{2}}{r}$

By rearranging this equation, we get,

$v=\sqrt{ar}\phantom{\rule{0ex}{0ex}}=\sqrt{3×{10}^{14}m/{s}^{2}×0.15m}\phantom{\rule{0ex}{0ex}}=6.7×{10}^{6}m/s$

Therefore, the speed of electron is$6.7×{10}^{6}m/s$ .

## Step 4: (b) Calculate the period of the motion

From equation (ii), we can write

$T=\frac{2\pi r}{V}$

Substitute the value of given radius and speed calculated in part (a), we get

$T=\frac{2\pi ×0.15m}{6.7×{10}^{6}m/s}\phantom{\rule{0ex}{0ex}}=1.4×{10}^{-7}s$

Therefore, the time period of rotation of electron is $1.4×{10}^{-7}s$.

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