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
