Respuesta :
Answer:
Explanation:
Given that,
Charge q=-5.90nC
Magnetic field B= -1.28T k
And the magnetic force
F =−( 3.70×10−7N )i+( 7.60×10−7N )j
Let the velocity be V(xi + yj + zk)
Then, the force is given as
Note i×i=j×j×k×k=0
i×j=k. j×i=-k
j×k=i. k×j=-i
k×i=j. i×k=-j
The force in a magnetic field is given as
F= q(v×B)
−( 3.70×10−7N )i+( 7.60×10−7N )j =
q(xi + yj + zk) × -1.28k
−( 3.70×10−7N )i+( 7.60×10−7N )j=
q( -1.28x i×k - 1.28y j×k - 1.28z k×k)
−( 3.70×10−7N )i+( 7.60×10−7N )j=
q( 1.28xj - 1.28y i )
−( 3.70×10−7N )i+( 7.60×10−7N )j=
q( -1.28y i + 1.28x j)
So comparing comparing coefficients
let compare x axis component
-( 3.70×10−7N )i=-1.28qy i
−3.70×10−7N = -1.28qy
y= -3.7×10^-7/-1.28q
y= -3.7×10^-7/-1.28×-5.90×10^-9)
y=-48.99m/s
y≈-49m/s
Let compare y-axisaxis
7.6×10−7N j = 1.28qx j
7.6×10−7N = 1.28qx
x= 7.6×10^-7/1.28q
x= 7.6×10^-7/1.28×-5.90×10^-9
x=-100.64m/s
a. Then, the velocity of the x component is x= -100.64m/s
b. Also, the velocity component of the y axis is y=-49m/s
c. We will compute
V•F
V=-100.64i -49j
F=−( 3.70×10−7 N )i+( 7.60×10−7 N )j
Note
i.j=j.i=0. Also i.i = j.j =1
V•F is
(-100.64i-49j)•−(3.70×10−7N)i+(7.60×10−7 N )j =
3.724×10^-5 - 3.724×10^-5=0
V•F=0
d. Angle between V and F
V•F=|V||F|Cosx
0=|V||F|Cosx
Cosx=0
x= arccos(0)
x=90°
Since the dot product is zero, from vectors , if the dot product of two vectors is zero, then the vectors are perpendicular to each other
The magnetic force and the scalar product allows to find the results for the questions about the movement of the particle in the field of magnetism are:
a) The velocities are: vₓ = -100.6 m / s, [tex]v_y[/tex] = 49.0 m / s
b) The z-axis component of velocity cannot be determined.
c) Velocity and force are perpendicular.
Given parameters
- The electric charge of the particle is Q = 5.960 nC = 5.9 10-9 C
- The magnetic field B = -1.25 k ^
- The magnetic force Fm = (-3.70 i ^ + 7.60 j ^) 107 N
To find
a) The speeds
b) there are components that cannot be determined
c) The scalar product of v.F and the angle
The magnetic force is given by the vector product between the speed and the magnetic field.
F = q v x B
Where the bold letters indicate vectors, F is the force, q the charge, v the velocity and B the magnetic field.
The best way to find forces is by solving the determinant.
[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]
a) Let's write the expression for the magnetic force.
[tex]F_x i + F_y j + F_z k = q \ \left[\begin{array}{ccc}i&j&k\\v_x&v_y&v_z\\0&0&B_z\end{array}\right][/tex]
[tex]F_x i + F_y j + F_z k = q \ (i v_y B_z - j v_x B_z)[/tex]
let's write the components on each axis.
[tex]F_x = q v_yB_z \\F_y = q v_x B_z[/tex]
Let's calculate.
-3.70 10⁻⁷ = 5.90 10⁻⁹ [tex]v_y[/tex] (-1.28)
3.70 10² = 7.552 [tex]v_y[/tex]
[tex]v_y = \frac{3.70 \ 10^2 }{7.552}[/tex]
[tex]v_y[/tex] = 49.0 m / s
7.60 10⁻⁷ = 5.90 10⁻⁹ vₓ (-1.28)
7.60 10² = -7.552 vₓ
vₓ = [tex]- \frac{7.60 \ 10^2 }{7.552}[/tex]
vₓ = - 1.006 10²
vₓ = -100.6 m / s
b) The component on the z axis cannot be determined from the given information since the vector product is zero
c) ask for the dot product and the angle between the vectors
v. F = ([tex]v_x i +v_y j + v_z k[/tex]) . ( [tex]F_x i + F_y j + 0 k[/tex] )
v.F = [tex]v_x F_x + v_yF_y[/tex]
v .F = - 100.6 (-3.70 10⁻⁷) + 49.0 7.60 10⁻⁷
v.F = -3.72 105 + 3.72 105
v . F = 0
The scale product can be written
v. F = |v| |F| cos θ
v. F = 0
Therefore, the cos θ = 0, this occurs for an angle of θ = 90º
Therefore the velocity and the forces are perpendicular.
In conclusion, using the magnetic force and the scalar product we can find the results for the questions about the movement of the particle in the field of magnetism are:
a) The velocities are: vₓ = -100.6 m / s, [tex]v_y[/tex] = 49.0 m / s
b) The z-axis component of velocity cannot be determined.
c) Velocity and force are perpendicular, the angle is: θ = 90º.
Learn more here: brainly.com/question/13791875
