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Arada [10]
3 years ago
12

A particle of charge q moving through a magnetic field B experiences the force where v is the velocity of the particle. Show tha

t: (i) The force does no work in moving the particle along its path from t = 0 to t=T. (i1) The speed v(t) Iv(O)ll of the particle is constant in time e particle 1s constant in time
Physics
1 answer:
MrRa [10]3 years ago
3 0

Answer:

       dx/Dt x B . x  =0

Explanation:

Let's calculate the work and the magnetic force, the expression for magnetic force is

        F = qv x B

Bold indicate vector quantities, the expression for the job is

         W = F. X

Let's replace in this equation

       W = q v x B . X

The definition of speed is

      v =  dX / dt

With what work is left

     W = q dX / dt x B . X

As we can see the vector product gives us a vector perpendicular to dX and its scalar product by X of zero

Second part

   The speed a vector and although the magnitude is constant the change of direction implies a change in the speed.

   Let's calculate the magnitudes of speed (speed)

       F = qv B sin θ

       F = ma

       q v B sin θ = ma

       a = qvB / m senT

     

This acceleration is perpendicular to the magnetic field and the velocity, so it does not change if magnitude but its direction, it is directed to the center of the circle.

   | v | = q vB/m sin θ

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I need help with this physics question
insens350 [35]
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The question is trying to get you to realize that to get from a reference point to a certain position, you have to know

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4 0
3 years ago
A 7.94-nC charge is located 1.77 m from a 4.14-nC point charge. (a) Find the magnitude of the electrostatic force that one charg
STatiana [176]

Answer:

F=94.32*10⁻⁹N  , The force F is repusilve because both charges have the same sign (+)

Explanation:

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:  

F=K*q₁*q₂/d² Formula (1)  

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁,q₂:Charges in Coulombs (C)  

d: distance between the charges in meters(m)

Equivalence  

1nC= 10⁻⁹C

Data

K=8.99x10⁹N*m²/C²

q₁ = 7.94-nC= 7.94*10⁻⁹C

q₂= 4.14-nC=  4.14 *10⁻⁹C

d= 1.77 m

Magnitude of the electrostatic force that one charge exerts on the other

We apply formula (1):

F=8.99x10^{9} *\frac{7.94*10^{-9} *4.14 *10^{-9} }{1.77^{2} }

F=94.32*10⁻⁹N  , The force F is repusilve because both charges have the same sign (+)

7 0
3 years ago
45. A particle moves in a straight line with an initial velocity of 30 m/s and constant acceleration 30 m/s2 . (a) What is its d
omeli [17]

Answer:

Displacement after 5 seconds is 155/2 meters

Explanation:

Let X (t) represent the equation of the position, then you have to d2x / dt2 = 5.

Applying the fundamental theorem of the calculation dx/dt = 5t + vo. The speed equation is V (t) = 5t + vo. Since the initial velocity is 30m/s, V (0) = 5 (0) + vo = 30. Therefore, V (t) = dx/dt = 5t + 30. Applying again the fundamental theorem of the calculation X (t) = 5t^2 / 2 + 30t + xo.

Displacement in 5 seconds is given by X (5) - X (0).

X (5) - X (0) = 5 (5)^2/2 +3 (5) + Xo - 5 (0)^2/2 -3 (0) -Xo = 155/2

Displacement after 5 seconds is 155/2 meters

7 0
3 years ago
Read 2 more answers
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