Answer:
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
