Question

Question #b27b7

Answer 1

`"a)"` `1.25 times 10^(- 13)` `"N"`

`"b)"` `2.89 times 10^(- 2)` `"N/m"`

Explanation:

`"1"`

`"a)"`

We are given two vectors, one for the velocity `(vec(v))` and one for the magnetic field `(vec(B))`.

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

`Rightarrow |vec(v)| = sqrt((4.0 times 10^(6))^(2) + (6.0 times 10^(6))^(2))`

`Rightarrow |vec(v)| = sqrt(52,000,000,000,000)`

`therefore |vec(v)| = 7,211,102.551`

`and`

`Rightarrow |vec(B)| = sqrt(0.030^(2) + (- 0.15)^(2))`

`Rightarrow |vec(B)| = sqrt(0.0234)`

`therefore |vec(B)| = 0.1529705854`

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

`Rightarrow cos(theta) = frac(vec(v) cdot vec(B))(|vec(v)| cdot |vec(B)|)`

`Rightarrow cos(theta) = frac((4.0 times 10^(6) times 0.030) + (6.0 times 10^(6) times (- 0.15)))(7,211,102.551 cdot 0.1529705854)`

`Rightarrow cos(theta) = frac(120,000 + (- 900,000))(1,103,086.5786)`

`Rightarrow cos(theta) = frac(- 780,000)(1,103,086.5786)`

`Rightarrow cos(theta) = - 0.7071067812`

`Rightarrow theta = 2.35619449`

`therefore theta = 135^(circ)`

Now, let's use the formula `F = q v B sin(theta)`; 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:

`Rightarrow F = (1.6 times 10^(- 19)) cdot (7,211,102.551) cdot (0.1529705854) cdot sin(135^(circ))`

`Rightarrow F = 1.6 times 10^(- 19) cdot 1,103,086.5786 cdot 0.70710678118`

`Rightarrow 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)` `"N"`.

`"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 `(4 pi times 10^(- 7))`, `I` is the current, and `d` is the distance between the two wires:

`Rightarrow frac(F)(L) = frac(4 cdot pi times 10^(- 7) cdot 17^(2))(2 cdot pi cdot 0.002)`

`Rightarrow frac(F)(L) = frac(2 times 10^(- 7) cdot 289)(0.002)`

`Rightarrow 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)` `"N/m"`.

Answer 2

a: `F=0.42*10^6veck`
b: `F/l=0.0289`

Explanation:

Let us first solve for (a)

we should know some vector properties

for this problem ,

`veci xxvecj=veck`

`vecjxxveci=-veck`
`vecixxveci=0`
`vecjxxvecj=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=qE+q*(vecvxxvecB)`

since there is no electric field `E=0`
Hence `F=q*(vecvxxvec B)`

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 `color(red)vecv` and `color(red)vecB`

`color(red)(vecvxxvecB)=(4.0*10^6veci+6.0*10^6vecj)xx(0.030veci+0.15vecj)`

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)#

`color(red)(vecvxxvecB)=0+(6.0*10^6*0.030)(-veck)+(4.0*10^6*0.15)(veck)+0`

On summing up we get `color(red)(vecvxxvecB)=(0.6*10^6-0.18*10^6)veck`

Hence the magnetic force on the proton is given by

`F=1*(0.42*10^6)veck`
`color(blue)(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=(muI_1I_2l)/(2pid)`
where

`mu` is the permeability of free space which is equal to `4pi*10^-7`newton/ `ampere^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
`F/l=(mu*I_1*I_2)/(2pi*d)`

substitute the values to get the answer

`I_1=I_2=17A`
`d=2*10^-3m`
`color(red)(F/l)=(4pi*10^-7*17*17)/(2*pi*2*10^-3`

`color(red)(F/l)=17^2*10^-4`
`color(red)(F/l)=0.0289` N/m