A particle with a charge of − 5.10 nC is moving in a uniform magnetic field of B⃗ =−( 1.20 T )k^. The magnetic force on the particle is measured to be F⃗ =−( 3.30×10−7 N )i^+( 7.60×10−7 N )j^.

A) Calculate the x-component of the velocity of the particle
B) Calculate the y-component of the velocity of the particle
C) Calculate the scalar product v⃗ ⋅F
D) What is the angle between v⃗ and F⃗ ? Give your answer in degrees.

The x-component of the velocity of the particle is -124.2 m/s.

The y-component of the velocity of the particle is -53.92 m/s.

The scalar (dot) product of velocity (v) and magnetic force  (F) is 0

The angle between velocity (v) and magnetic force  (F) is 90⁰.

The given parameters;

The magnitude of the magnetic force at given field and velocity is calculated as follows;

"this is read as the product of q and v cross B."

The y-component of the velocity of the particle is calculated as;

The x-component of the velocity of the particle is calculated as;

The dot product of v and F is calculated as;

The angle between v and F is calculated as follows;

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Explanation:

Given that,

Charge q=-5.10nC

Magnetic field B= -1.2T k

And the magnetic force

F =−( 3.30×10−7N )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

F= q(v×B)

−( 3.30×10−7N )i+( 7.60×10−7N )j =

q(xi + yj + zk) × -1.2k

−( 3.30×10−7N )i+( 7.60×10−7N )j=

q( -1.2x i×k - 1.2y j×k - 1.2z k×k)

−( 3.30×10−7N )i+( 7.60×10−7N )j=

q( 1.2xj - 1.2y i )

−( 3.30×10−7N )i+( 7.60×10−7N )j=

q( -1.2y i + 1.2x j)

So comparing comparing coefficients

let compare x axis component

-( 3.30×10−7N )i=-1.2qy i

−3.30×10−7N = -1.2qy

y= -3.3×10^-7/-1.2q

y= -3.3×10^-7/-1.2×-5.10×10^-9)

y=-53.92m/s

Let compare y-axisaxis

7.6×10−7N j = 1.2qx j

7.6×10−7N = 1.2qx

x= 7.6×10^-7/-1.2q

x= 7.6×10^-7/1.2×-5.10×10^-9)

x=-124.18m/s

a. Then, the velocity of the x component is x= -124.18m/s

b. Also, the velocity component of the y axis is =-53.92m/s

c. We will compute

V•F

V=-124.18i -53.92j

F=−( 3.30×10−7 N )i+( 7.60×10−7 N )j

Note

i.j=j.i=0. Also i.i = j.j =1

V•F is

(-124.18i-53.92j)•−(3.30×10−7N)i+(7.60×10−7 N )j =

4.1×10^-5 - 4.1×10^-5=0

V•F=0

d. Angle between V and F

V•F=|V||F|Cosx

0=|V||F|Cos

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

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