November 15, 2019

Solution Set 9

Physics 208-02 – Introductory Physics II
Solution Set 9
Nazareth College
Department of Chemistry & Biochemistry
Robert F Szalapski, PhD – Adjunct Lecturer
Spring 2012

§ Chapter 20, Problem 9

Alpha particles of charge q=+2e and mass m=6.6\times 10^{{-27}}\,\mathrm{kg} are omitted from a radioactive source at a speed of v=1.6\times 10^{7}\,\mathrm{m}/\mathrm{s}. What magnetic field strength would be required to bend them in a circular path of radius r=0.25\;\mathrm{m}?

Assuming that the magnetic field is oriented perpendicular to the velocity vector of an alpha particle (so that \sin\theta=1), the magnetic force provides the centripetal acceleration.

\displaystyle qvB \displaystyle= \displaystyle\frac{mv^{2}}{r}
\displaystyle B \displaystyle= \displaystyle\frac{mv}{qr}
\displaystyle= \displaystyle\frac{mv}{2er}
\displaystyle= \displaystyle\frac{\left(6.6\times 10^{{-27}}\,\mathrm{kg}\right)\left(1.6\times 10^{7}\,\frac{\mathrm{m}}{\mathrm{s}}\right)}{2\left(1.602\times 10^{{-19}}\,\mathrm{C}\right)\left(0.25\;\mathrm{m}\right)}
\displaystyle= \displaystyle 1.3\;\mathrm{T}

§ Chapter 20, Problem 10

Determine the magnitude and direction of the force on an electron traveling 8.75\times 10^{5}\,\mathrm{m}/\mathrm{s} horizontally to the east in a vertically upward magnetic field of strength 0.75\;\mathrm{T}.

The magnitude is given by

\displaystyle F \displaystyle= \displaystyle qvB\sin\theta
\displaystyle= \displaystyle\left(1.602\times 10^{{-19}}\,\mathrm{C}\right)\left(8.75\times 10^{5}\,\frac{\mathrm{m}}{\mathrm{s}}\right)\left(0.75\;\mathrm{T}\right)\sin 90^{0}
\displaystyle= \displaystyle 1.1\times 10^{{-13}}\,\mathrm{N}

To find the direction, point the thumb of your right hand to the east and your fingers up, and the palm of your hand will be directed south. Because an electron has a negative charge, the direction of the force will be reversed. The direction is north.

§ Chapter 20, Problem 14

A 5.0\;\mathrm{MeV} (kinetic energy) proton enters a 0.20\;\mathrm{T} magnetic field in a plane perpendicular to the field (so that \sin\theta=1). What is the radius of its path?

With the magnetic force on the proton supplying the centripetal force,

\displaystyle qvB \displaystyle= \displaystyle\frac{mv^{2}}{r}
\displaystyle r \displaystyle= \displaystyle\frac{mv}{qB}
\displaystyle= \displaystyle\frac{m_{p}v}{eB}

We will first need to find the speed, v.

\displaystyle\frac{1}{2}m_{p}v^{2} \displaystyle= \displaystyle eV
\displaystyle v \displaystyle= \displaystyle\sqrt{\frac{2eV}{m_{p}}}
\displaystyle= \displaystyle\sqrt{\frac{2\left(1.602\times 10^{{-19}}\,\mathrm{C}\right)\left(5.0\times 10^{6}\,\mathrm{V}\right)}{1.67\times 10^{{-27}}\,\mathrm{kg}}}
\displaystyle= \displaystyle 3.1\times 10^{7}\,\frac{\mathrm{m}}{\mathrm{s}}

We may now solve for r.

\displaystyle r \displaystyle= \displaystyle\frac{\left(1.67\times 10^{{-27}}\,\mathrm{kg}\right)\left(3.1\times 10^{7}\,\frac{\mathrm{m}}{\mathrm{s}}\right)}{\left(1.602\times 10^{{-19}}\,\mathrm{C}\right)\left(0.20\;\mathrm{T}\right)}
\displaystyle= \displaystyle 1.6\;\mathrm{T}

§ Chapter 20, Problem 16

What is the velocity of a beam of electrons that go undeflected when passing through perpendicular electric and magnetic fields of magnitude E=8.8\times 10^{3}\,\mathrm{V}/\mathrm{m} and B=3.5\times 10^{{-3}}\,\mathrm{T}? What is the radius of the electron orbit if the electric field is turned off?

This arrangement is called a velocity selector. For it to work the electric field, the magnetic field and the velocity of the electron must all be mutually perpendicular. Assuming that it is set up correctly so that the electric force and the magnetic force oppose one another

\displaystyle qvB \displaystyle= \displaystyle qE
\displaystyle v \displaystyle= \displaystyle\frac{E}{B}
\displaystyle= \displaystyle\frac{8.8\times 10^{3}\,\frac{\mathrm{V}}{\mathrm{m}}}{3.5\times 10^{{-3}}\,\mathrm{T}}
\displaystyle= \displaystyle 2.5\times 10^{6}\,\frac{\mathrm{m}}{\mathrm{s}}

If the electric field is switched off it will move in a circle with the centripetal force supplied by the magnetic force.

\displaystyle qvB \displaystyle= \displaystyle\frac{mv^{2}}{r}
\displaystyle r \displaystyle= \displaystyle\frac{mv}{qB}
\displaystyle= \displaystyle\frac{m_{e}v}{eB}
\displaystyle= \displaystyle\frac{\left(9.11\times 10^{{-31}}\,\mathrm{kg}\right)\left(2.5\times 10^{6}\,\frac{\mathrm{m}}{\mathrm{s}}\right)}{\left(1.602\times 10^{{-19}}\,\mathrm{C}\right)\left(3.5\times 10^{{-3}}\,\mathrm{T}\right)}
\displaystyle= \displaystyle 4.1\times 10^{{-3}}\,\mathrm{m}

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