Respuesta :
Answer:
a
The value of magnetic field is [tex]B_a=6.56*10^{-19} T[/tex]
In the direction of the positive x - axis
b
The value of magnetic field is [tex]B_a=6.56*10^{-19} T[/tex]
In the direction of the positive z - axis
Explanation:
From the question we are told that
The velocity of the electron is [tex]v_z = 4.1*10^{7}m/s[/tex]
Considering the first position
The equation for the magnetic field is
[tex]\= B = \frac{\mu_0}{ 4 \pi} \frac{q \=v * \= r}{r^2}[/tex]
Now [tex]r = \sqrt{i^2 + j^2 + k^2}[/tex]
substituting values
[tex]r = \sqrt{1^2 +0^2 +0^2}[/tex]
[tex]= 1[/tex]
[tex]\= r = \frac{\r r}{r}[/tex]
[tex]\r r = 1 i + 0j + 0k[/tex]
Therefore [tex]\= r = \frac{1i + 0j + 0k}{1}[/tex]
[tex]= i[/tex]
So Substituting [tex]4 \pi *10^ {-7}[/tex] for [tex]\mu_o[/tex] , [tex]1.602 *10^{-19}[/tex] for q
[tex]\= B_a = \frac{4\pi *10^{-7}}{4 \pi} \frac{1.602 *10^{-19} * 4.1*10^{7} * i}{1^2}[/tex]
[tex]\= B_a=6.56*10^{-19} T (i)[/tex]
Considering the second position
Here
[tex]r = \sqrt{0^2 + 0^2 + 2^2}[/tex]
[tex]=2[/tex]
[tex]\= r = \frac{\r r}{r}[/tex]
[tex]\r r = 0 i + 0j + 2k[/tex]
[tex]\= r = \frac{0i + 0j + 2k}{2}[/tex]
[tex]= k[/tex]
So Substituting [tex]4 \pi *10^ {-7}[/tex] for [tex]\mu_o[/tex] , [tex]1.602 *10^{-19}[/tex] for q
[tex]\= B_a = \frac{4\pi *10^{-7}}{4 \pi} \frac{1.602 *10^{-19} * 4.1*10^{7} * k}{1^2}[/tex]
[tex]\= B_a=6.56*10^{-19} T (k)[/tex]
The Biot-Savart law and the expression for the magnetic force allow us to find the results for the questions about the magnetic field and the force at two different points are:
a) Point r = 1 i.
- The magnetic field is: 6.56 10⁻¹⁵ j T
- The magnetic force is: 4.3 10⁻²⁶ i N
b) Point r = 2 k
- The magnetic field is zero.
- The force is also zero.
The Biot-Savart law determines the value and direction of the magnetic field, using the relationship
B = [tex]\frac{\mu_o }{4\pi } I[/tex] ∫ [tex]\frac{1}{r^2 }[/tex] dl x r^
Where the bold letters indicate vectors, B is the magnetic field, μ₀ is the permeability of free space, I the current is the direction of the current and r the distance from the current to the point of interest.
In this case, we have an electron moving, let's transform the equation using that current is the change in charge per unit of time.
I dl = [tex]\frac{dq}{dt } \ dl = dq \ \frac{dl}{dt}[/tex]
The velocity is defined as the change in position per unit of time.
v= [tex]\frac{dl}{dt}[/tex]
Let's substitute
I dl = dq v
Let's substitute.
B = [tex]\frac{\mu_o }{4\pi } q[/tex] ∫ [tex]\frac{1}{r^2}[/tex] v x r^
We look for the distance between the position of the charge r₀ = 0 and the point of interest r = 1 i
r² = (x-x₀) ² + (y-y₀) ² + (z-z₀) ²
r² = (1-0)²
The unit vector of the position is:
r^ = 1 i + 0 j + 0k
We solve the vector product of the velocity by the unit vector of the displacement, for this we solve the determinant.
v x r^ = [tex]\left[\begin{array}{ccc}i&j&k\\v_x&v_y&v_z\\r_x&r_y&r_z\end{array}\right][/tex]
v x r^ = [tex]\left[\begin{array}{ccc}i&j&k\\0&0&4.1\\1&0&0\end{array}\right] 10^7 = - 4.1 \ 10^7 j[/tex]
We substitute into the Biot-Savart law to find the magnetic field.
B = [tex]\frac{4\pi \ 10^{-7} }{4\pi } (-1.6 10^{-19} ) \frac{(-4.1 \ 10^7 j) }{10^{-4}}[/tex]
B = 6.56 10⁻¹⁵ j^ T
The magnetic force is given by the vector product of the velocity and the magnetic field, therefore it is a vector quantity.
F = q v x B
Where the bold letters indicate vectors, F is the force, q the charge of the particle, v the velocity and B the magnetic field.
An easy way to find the force is to solve the determinant of the equation.
[tex]F = q \ \left[\begin{array}{ccc}i&j&k\\v_x&v_y&v_z\\B_x&B_y&B_z\end{array}\right][/tex]
Let's solve the determine to shorten the magnetic force.
[tex]F = -1.6 \ 10^{-19} \ \left[\begin{array}{ccc}i&j&k\\0&0&4.1\\0&6.56&0\end{array}\right] \ 10^{-8}[/tex]
F = 4.30 10⁻²⁶ i^ N
b) point r = 0 i + 0 j + 2 k cm
We look for the magnetic field.
the distance is
r² = 0 + 0 + (2-0) ²
r² = 4 10⁻⁴ m²
The unit vector in the direction.
r^ = (0 + 0 + 2 k) / 2
r^ = k
We use the Biot-Savart law to find the magnetic field.
B = [tex]\frac{4\pi \ 10^{-7} }{4\pi } (-1.6 \ 10^{-19}) \frac{1}{4 \ 10^{-4}}[/tex] vx r^
Let's solve the determinant
v x r = [tex]\left[\begin{array}{ccc}i&j&k\\0&0&4.1\\0&0&1\end{array}\right] \ 10^7[/tex]
v x r = 0
In this case, as the particle moving at the origin in the same direction of displacement, the field created is zero and therefore the magnetic force is zero.
In conclusion, using the Biot-Savart law and the expression for the magnetic force we can find the results for the questions about the magnetic field and the force at two different points are:
a) Point r = 1 i^
- The magnetic field is: 6.56 10-15 j^ T
- The magnetic force is: 4.3 10-26 i^ N
b) Point r = 2 k^
- The magnetic field is zero and therefore the force is also zero.
Learn more here: brainly.com/question/22629118
