A particle of mass \(m = 6.64 \times 10^{-27}\) kg with charge \(q = 3.2 \times 10^{-19}\) C moves with a velocity \(v = (2i + 3j - k) \times 10^5\) m/s through a region where both... A particle of mass \(m = 6.64 \times 10^{-27}\) kg with charge \(q = 3.2 \times 10^{-19}\) C moves with a velocity \(v = (2i + 3j - k) \times 10^5\) m/s through a region where both a uniform magnetic field and a uniform electric field can exist. The uniform magnetic field is \(B = (3i + 4j - 2k)\) T and the uniform electric field is not present initially. In this question, express all vectors in unit vector notation. (a) Calculate the magnetic force vector acting on the particle. (b) Find the radius of the circular motion. (c) Now the uniform electric field is switched on. The electric field vector is \(E = (4i - j - 2k) \times 10^5\) N/C. Find the net force vector acting on the particle. Read more

Steps to Solve

1. Calculate the magnetic force

The magnetic force on a charged particle is given by the formula: $$ F_B = q(v \times B) $$ where $q$ is the charge, $v$ is the velocity vector, and $B$ is the magnetic field vector. We need to compute the cross product $v \times B$.

Given $v = (2i+3j-k)\times10^5$ m/s and $B = (3i + 4j-2k)$ T, we calculate the cross product as follows:

$$ v \times B = \begin{vmatrix} i & j & k \ 2\times10^5 & 3\times10^5 & -1\times10^5 \ 3 & 4 & -2 \end{vmatrix} $$

$$ v \times B = i[(3\times10^5)(-2) - (-1\times10^5)(4)] - j[(2\times10^5)(-2) - (-1\times10^5)(3)] + k[(2\times10^5)(4) - (3\times10^5)(3)] $$

$$ v \times B = i[-6\times10^5 + 4\times10^5] - j[-4\times10^5 + 3\times10^5] + k[8\times10^5 - 9\times10^5] $$

$$ v \times B = -2\times10^5 i + 1\times10^5 j - 1\times10^5 k $$

Now, multiply by the charge $q = 3.2\times10^{-19}$ C:

$$ F_B = (3.2\times10^{-19}) (-2\times10^5 i + 1\times10^5 j - 1\times10^5 k) $$

$$ F_B = (-6.4i + 3.2j - 3.2k)\times10^{-14} \text{ N} $$

2. Find the radius of the circular motion

The magnetic force causes the particle to move in a circular path. The magnetic force provides the centripetal force, so $F_B = \frac{mv^2}{r}$, where $r$ is the radius of the circle. However, since the velocity is not perpendicular to the magnetic field, we need to find the component of velocity perpendicular to the magnetic field, $v_{\perp}$. And we will use $F_B = q v_{\perp} B$, where $B$ is the magnitude of the magnetic field. Then, $qv_{\perp}B = \frac{mv_{\perp}^2}{r}$, therefore $r = \frac{mv_{\perp}}{qB}$.

First, we find the magnitude of the velocity: $|v| = \sqrt{(2\times10^5)^2 + (3\times10^5)^2 + (-1\times10^5)^2} = \sqrt{4+9+1}\times10^5 = \sqrt{14}\times10^5$ m/s Then, we find the magnitude of the magnetic field: $|B| = \sqrt{3^2 + 4^2 + (-2)^2} = \sqrt{9+16+4} = \sqrt{29}$ T Then, we use the magnetic force magnitude that we previously calculated: $|F_B| = \sqrt{(-6.4\times10^{-14})^2 + (3.2\times10^{-14})^2 + (-3.2\times10^{-14})^2} = \sqrt{40.96+10.24+10.24}\times10^{-14} = \sqrt{61.44}\times10^{-14} \approx 7.84\times10^{-14}$ N

We have $|F_B|=q v_{\perp} B$, so $v_{\perp} = \frac{|F_B|}{q|B|} = \frac{7.84\times10^{-14}}{(3.2\times10^{-19})(\sqrt{29})} = \frac{7.84\times10^5}{3.2\sqrt{29}} \approx 4.55\times10^4$ m/s

Now we can calculate the radius: $r = \frac{mv_{\perp}}{qB} = \frac{(6.64\times10^{-27})(4.55\times10^4)}{(3.2\times10^{-19})(\sqrt{29})} = \frac{3.02\times10^{-22}}{1.72\times10^{-18}} \approx 1.75\times10^{-4}$ m

3. Calculate the net force with the electric field

The electric force on the particle is given by: $$ F_E = qE $$ where $E = (4i-j-2k)\times10^5$ N/C. So, $$ F_E = (3.2\times10^{-19}) (4i-j-2k)\times10^5 = (12.8i - 3.2j - 6.4k)\times10^{-14} \text{ N} $$

The net force is the sum of the magnetic and electric forces: $$ F_{net} = F_B + F_E = (-6.4i + 3.2j - 3.2k)\times10^{-14} + (12.8i - 3.2j - 6.4k)\times10^{-14} $$

$$ F_{net} = (12.8 - 6.4)i + (3.2 - 3.2)j + (-3.2 - 6.4)k \times10^{-14} $$

$$ F_{net} = (6.4i + 0j - 9.6k)\times10^{-14} \text{ N} $$