# Question b27b7

Sep 9, 2017

$\text{a)}$ $1.25 \times {10}^{- 13}$ $\text{N}$

$\text{b)}$ $2.89 \times {10}^{- 2}$ $\text{N/m}$

#### Explanation:

$\text{1}$

$\text{a)}$

We are given two vectors, one for the velocity $\left(\vec{v}\right)$ and one for the magnetic field $\left(\vec{B}\right)$.

First, let's evaluate the magnitudes of both vectors:

$R i g h t a r r o w | \vec{v} | = \sqrt{{\left(4.0 \times {10}^{6}\right)}^{2} + {\left(6.0 \times {10}^{6}\right)}^{2}}$

$R i g h t a r r o w | \vec{v} | = \sqrt{52 , 000 , 000 , 000 , 000}$

$\therefore | \vec{v} | = 7 , 211 , 102.551$

$\mathmr{and}$

$R i g h t a r r o w | \vec{B} | = \sqrt{{0.030}^{2} + {\left(- 0.15\right)}^{2}}$

$R i g h t a r r o w | \vec{B} | = \sqrt{0.0234}$

$\therefore | \vec{B} | = 0.1529705854$

Then, let's find the angle between these two vectors:

$R i g h t a r r o w \cos \left(\theta\right) = \frac{\vec{v} \cdot \vec{B}}{| \vec{v} | \cdot | \vec{B} |}$

$R i g h t a r r o w \cos \left(\theta\right) = \frac{\left(4.0 \times {10}^{6} \times 0.030\right) + \left(6.0 \times {10}^{6} \times \left(- 0.15\right)\right)}{7 , 211 , 102.551 \cdot 0.1529705854}$

$R i g h t a r r o w \cos \left(\theta\right) = \frac{120 , 000 + \left(- 900 , 000\right)}{1 , 103 , 086.5786}$

$R i g h t a r r o w \cos \left(\theta\right) = \frac{- 780 , 000}{1 , 103 , 086.5786}$

$R i g h t a r r o w \cos \left(\theta\right) = - 0.7071067812$

$R i g h t a r r o w \theta = 2.35619449$

$\therefore \theta = {135}^{\circ}$

Now, let's use the formula $F = q v B \sin \left(\theta\right)$; where $q$ is the charge of the proton, $v$ is it's velocity, and $\theta$ is the angle between the velocity and magnetic field vectors:

$R i g h t a r r o w F = \left(1.6 \times {10}^{- 19}\right) \cdot \left(7 , 211 , 102.551\right) \cdot \left(0.1529705854\right) \cdot \sin \left({135}^{\circ}\right)$

$R i g h t a r r o w F = 1.6 \times {10}^{- 19} \cdot 1 , 103 , 086.5786 \cdot 0.70710678118$

$R i g h t a r r o w F = 1.6 \times {10}^{- 19} \cdot 780 , 000$

$\therefore F = 1.248 \times {10}^{- 13}$

Therefore, the force on the proton is around $1.25 \times {10}^{- 13}$ $\text{N}$.

$\text{b)}$

For this question, let's use the equation $\frac{F}{L} = \frac{{\mu}_{0} {I}^{2}}{2 \pi d}$; where $\frac{F}{L}$ is the force per unit length, ${\mu}_{0}$ is the magnetic constant $\left(4 \pi \times {10}^{- 7}\right)$, $I$ is the current, and $d$ is the distance between the two wires:

$R i g h t a r r o w \frac{F}{L} = \frac{4 \cdot \pi \times {10}^{- 7} \cdot {17}^{2}}{2 \cdot \pi \cdot 0.002}$

$R i g h t a r r o w \frac{F}{L} = \frac{2 \times {10}^{- 7} \cdot 289}{0.002}$

$R i g h t a r r o w \frac{F}{L} = 0.0001 \cdot 289$

$\therefore \frac{F}{L} = 0.0289$

Therefore, the force per unit length between the wires is $2.89 \times {10}^{- 2}$ $\text{N/m}$.

Sep 9, 2017

a: $F = 0.42 \cdot {10}^{6} \vec{k}$
b: $\frac{F}{l} = 0.0289$

#### Explanation:

Let us first solve for (a)

we should know some vector properties

for this problem ,

$\vec{i} \times \vec{j} = \vec{k}$

$\vec{j} \times \vec{i} = - \vec{k}$
$\vec{i} \times \vec{i} = 0$
$\vec{j} \times \vec{j} = 0$

Okay Let's move on to the problem
Here it is given a magnetic field in x and y direction and also the velocity in in x and y direction

we use the relation of lorentz force
which is
$F = q E + q \cdot \left(\vec{v} \times \vec{B}\right)$

since there is no electric field $E = 0$
Hence $F = q \cdot \left(\vec{v} \times \vec{B}\right)$

Where ,
q is the charge of the particle,
v is the velocity of the particle and
B is the magnetic field

Now just vector multiplication of $\textcolor{red}{\vec{v}}$ and $\textcolor{red}{\vec{B}}$

$\textcolor{red}{\vec{v} \times \vec{B}} = \left(4.0 \cdot {10}^{6} \vec{i} + 6.0 \cdot {10}^{6} \vec{j}\right) \times \left(0.030 \vec{i} + 0.15 \vec{j}\right)$

Expanding the equation we get  color (red)(vec vxxvecB)=(4.0*10^6*0.030)(vecixxvecj)+(6.0*10^6*0.030)(vecjxxveci)+(4.0*10^6*0.15)(vecixxvecj)+(6.0*10^6*0.15)(vecjxxvecj)

$\textcolor{red}{\vec{v} \times \vec{B}} = 0 + \left(6.0 \cdot {10}^{6} \cdot 0.030\right) \left(- \vec{k}\right) + \left(4.0 \cdot {10}^{6} \cdot 0.15\right) \left(\vec{k}\right) + 0$

On summing up we get $\textcolor{red}{\vec{v} \times \vec{B}} = \left(0.6 \cdot {10}^{6} - 0.18 \cdot {10}^{6}\right) \vec{k}$

Hence the magnetic force on the proton is given by

$F = 1 \cdot \left(0.42 \cdot {10}^{6}\right) \vec{k}$
$\textcolor{b l u e}{q = 1}$ Since the charge of the proton is one

Hence the answer will be color(green)(F=0.42*10^6veck

Okay let's move on to (b)

$$Force between  two parallel current carrying wires is given by


$F = \frac{\mu {I}_{1} {I}_{2} l}{2 \pi d}$
where

$\mu$ is the permeability of free space which is equal to $4 \pi \cdot {10}^{-} 7$newton/ $a m p e r {e}^{2}$

${I}_{1}$ is the current in first wire
${I}_{2}$ is the current in the second wire
$l$ is the total length of the wires
$d$ is the distance between the two wires

Then the force per unit length is given by
$\frac{F}{l} = \frac{\mu \cdot {I}_{1} \cdot {I}_{2}}{2 \pi \cdot d}$

substitute the values to get the answer

${I}_{1} = {I}_{2} = 17 A$
$d = 2 \cdot {10}^{-} 3 m$
color(red)(F/l)=(4pi*10^-7*17*17)/(2*pi*2*10^-3#

$\textcolor{red}{\frac{F}{l}} = {17}^{2} \cdot {10}^{-} 4$
$\textcolor{red}{\frac{F}{l}} = 0.0289$ N/m