Physics, 23.03.2020 18:20, aorilneedshelp6636

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.

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answered: paulitaaustin

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;

charge on the particle, q = -5.10 nCmagnetic field strength, B = -(1.2 T)kmagnetic force on the particle, F = -(3.3 x 10⁻⁷)i + (7.6 x 10⁻⁷)j

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

$F = q(v \times B)$

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

$(-3.3\times 10^{-7})i + (7.6\times 10^{-7})j = q(v_x_i + v_y_j + v_z_k)\times (-1.2k)\\\\(3.3\times 10^{-7})i + (7.6\times 10^{-7})j = -1.2qv_x(i\times k) -1.2qv_y(j\times k) \\\\-1.2qv_z(k\times k)\\\\(-3.3\times 10^{-7})i + (7.6\times 10^{-7})j = -1.2qv_x(-j) - 1.2qv_y(i)\\\\(-3.3\times 10^{-7})i + (7.6\times 10^{-7})j = 1.2qv_x(j) - 1.2qv_y(i)$

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

$(-3.3\times 10^{-7})i = -1.2qv_y(i)\\\\v_y = \frac{-3.3\times 10^{-7}}{-1.2\times (-5.1 \times 10^{-9})} = - 53.92 \times 10^{12} \ m/s$

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

$(7.6 \times 10^{-7})j = 1.2qv_x(j)\\\\v_x = \frac{7.6\times 10^{-7}}{1.2 \times (-5.1 \times 10^{-9})} = -124.2 \ m/s$

The dot product of v and F is calculated as;

$V.F = (-124.2 i\ - \ 53.92j)\ .\ (-3.3\times 10^{-7}i \ + \ 7.6 \times 10^{-7})\\\\V.F = (4.098 \times 10^{-5})(i^2) \ - \ 4.098 \times 10^{-5})(j^2)\\\\V.F = (4.098 \times 10^{-5}) - (4.098 \times 10^{-5})\\\\V.F = 0$

The angle between v and F is calculated as follows;

$cos(\theta) = \frac{V.F}{|V||F|} \\\\cos(\theta) = \frac{0}{|V||F|} \\\\cos(\theta) = 0\\\\\theta = cos^{-1} (0)\\\\\theta = 90 ^0$

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answered: chuncky812

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

answered: Guest

hi,

what occurs during a solar eclipse?

there are small tides across earth.

the moon is between earth and the sun.

p.s. im not sure if the first answer is correct because during a solar ecliphse there are spring tides and i dont know if they are the same.

answered: Guest

The object's initial distance above the groundthe object's initial velocity

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A particle with a charge of − 5.10 nC is moving in a uniform magnetic field of B⃗ =−( 1.20 T )k^. Th...