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.

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