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Q102P

Expert-verifiedFound in: Page 91

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 ${\mathbf{3}}{\mathbf{.}}{\mathbf{0}}{\mathbf{\times}}{{\mathbf{10}}}^{{\mathbf{14}}}{\mathbf{m}}{\mathbf{/}}{{\mathbf{s}}}^{{\mathbf{2}}}$**** 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 $6.7\times {10}^{6}m/s$.

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

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

2) Radius of the circular path,

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

**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)

From equation (i), we can write,

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

By rearranging this equation, we get,

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

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

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

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

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

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