Question

in the figure, a particle moves along a circle in a region of uniform magnetic field of magnitude b = 4.9 mt. the particle is either a proton or an electron (you must decide which). it experiences a magnetic force of magnitude 3.2×10^(-15) n. what are (a) the particles speed, (b) the radius of the circle, and (c) the period of the motion? (a) number units (b) number units (c) number units

Explanation:

Step1: Identify the formula for magnetic - force on a charged particle in circular motion

The magnetic force $F = qvB$ provides the centripetal force $F_c= \frac{mv^{2}}{r}$, so $qvB=\frac{mv^{2}}{r}$, and $v=\frac{qBr}{m}$. Also, the period $T=\frac{2\pi r}{v}=\frac{2\pi m}{qB}$. The charge of a proton $q_p = 1.6\times10^{- 19}\ C$, mass of a proton $m_p=1.67\times10^{-27}\ kg$, charge of an electron $q_e=- 1.6\times10^{-19}\ C$, mass of an electron $m_e = 9.11\times10^{-31}\ kg$, and $B = 4.9\times10^{-3}\ T$, $F = 3.2\times10^{-15}\ N$. Since $F = qvB$, then $v=\frac{F}{qB}$.

Step2: Determine the particle

Since $F = qvB$, and $v=\frac{F}{qB}$. The magnitude of the charge of a proton and an electron is the same $q = 1.6\times10^{-19}\ C$. The mass of a proton is much larger than that of an electron. From $F = qvB$, we can find the speed. First, assume the particle is a proton or an electron. Given $F = 3.2\times10^{-15}\ N$ and $B = 4.9\times10^{-3}\ T$ and $q = 1.6\times10^{-19}\ C$. Then $v=\frac{F}{qB}=\frac{3.2\times10^{-15}}{1.6\times10^{-19}\times4.9\times10^{-3}}=\frac{3.2\times10^{-15}}{7.84\times10^{-22}}\approx4.08\times10^{6}\ m/s$.

Step3: Calculate the radius

From $qvB=\frac{mv^{2}}{r}$, we can get $r=\frac{mv}{qB}$. If it is a proton, $m = m_p=1.67\times10^{-27}\ kg$, $q = 1.6\times10^{-19}\ C$, $v\approx4.08\times10^{6}\ m/s$, $B = 4.9\times10^{-3}\ T$. Then $r=\frac{1.67\times10^{-27}\times4.08\times10^{6}}{1.6\times10^{-19}\times4.9\times10^{-3}}=\frac{6.8236\times10^{-21}}{7.84\times10^{-22}}\approx8.7\ m$.

Step4: Calculate the period

Using $T=\frac{2\pi m}{qB}$, for a proton, $m = m_p=1.67\times10^{-27}\ kg$, $q = 1.6\times10^{-19}\ C$, $B = 4.9\times10^{-3}\ T$. Then $T=\frac{2\pi\times1.67\times10^{-27}}{1.6\times10^{-19}\times4.9\times10^{-3}}=\frac{1.05\times10^{-26}}{7.84\times10^{-22}}\approx1.34\times10^{-5}\ s$.

Answer:

(a) Speed: $4.08\times10^{6}\ m/s$
(b) Radius (assuming proton): $8.7\ m$
(c) Period (assuming proton): $1.34\times10^{-5}\ s$
Units:
(a) $m/s$
(b) $m$
(c) $s$