A particle with charge −− 5.90 nCnC is moving in a uniform magnetic field B⃗ =−(B→=−( 1.28 TT )k^k^. The magnetic force on the particle is measured to be F⃗ =−(F→=−( 3.70×10−7 NN )i^+()i^+( 7.60×10−7 NN )j^)j^.

(a) Calculate all the components of the velocity of the particle that you can from this information.
(b) Are there components of the velocity that are not determined by the measurement of the force? Explain.
(c) Calculate the scalar product v⃗ ⋅F⃗ . What is the angle between v⃗ and F⃗

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, = 49.0 m / s

     b) The z-axis component of velocity cannot be determined.

     c) Velocity and force are perpendicular.

Given parameters

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.

a) Let's write the expression for the magnetic force.

let's write the components on each axis.

Let's calculate.

      -3.70 10⁻⁷ = 5.90 10⁻⁹  (-1.28)

      3.70 10² = 7.552  

      = 49.0 m / s

      7.60 10⁻⁷ = 5.90 10⁻⁹ vₓ (-1.28)

      7.60 10² = -7.552 vₓ

       vₓ =  

       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 = () . ( )

          v.F =    

          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,  = 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.

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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