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
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;
[tex]F = q(v \times B)[/tex]
"this is read as the product of q and v cross B."
[tex](-3.3\times 10^{-7})i + (7.6\times 10^{-7})j = q(v_x_i + v_y_j + v_z_k)\times (-1.2k)\\\\(3.3\times 10^{-7})i + (7.6\times 10^{-7})j = -1.2qv_x(i\times k) -1.2qv_y(j\times k) \\\\-1.2qv_z(k\times k)\\\\(-3.3\times 10^{-7})i + (7.6\times 10^{-7})j = -1.2qv_x(-j) - 1.2qv_y(i)\\\\(-3.3\times 10^{-7})i + (7.6\times 10^{-7})j = 1.2qv_x(j) - 1.2qv_y(i)[/tex]
The y-component of the velocity of the particle is calculated as;
[tex](-3.3\times 10^{-7})i = -1.2qv_y(i)\\\\v_y = \frac{-3.3\times 10^{-7}}{-1.2\times (-5.1 \times 10^{-9})} = - 53.92 \times 10^{12} \ m/s[/tex]
The x-component of the velocity of the particle is calculated as;
[tex](7.6 \times 10^{-7})j = 1.2qv_x(j)\\\\v_x = \frac{7.6\times 10^{-7}}{1.2 \times (-5.1 \times 10^{-9})} = -124.2 \ m/s[/tex]
The dot product of v and F is calculated as;
[tex]V.F = (-124.2 i\ - \ 53.92j)\ .\ (-3.3\times 10^{-7}i \ + \ 7.6 \times 10^{-7})\\\\V.F = (4.098 \times 10^{-5})(i^2) \ - \ 4.098 \times 10^{-5})(j^2)\\\\V.F = (4.098 \times 10^{-5}) - (4.098 \times 10^{-5})\\\\V.F = 0[/tex]
The angle between v and F is calculated as follows;
[tex]cos(\theta) = \frac{V.F}{|V||F|} \\\\cos(\theta) = \frac{0}{|V||F|} \\\\cos(\theta) = 0\\\\\theta = cos^{-1} (0)\\\\\theta = 90 ^0[/tex]
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