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Sergeeva-Olga [200]
3 years ago
14

A charge enters a uniform magnetic field oriented perpendicular to its path. The field deflects the particle into a circular arc

of radius R. If the charged particle enters the same magnetic field with three times the speed, what will be the radius of the circular arc?
Physics
1 answer:
frozen [14]3 years ago
7 0

Answer:

The new radius is three times the initial radius

Explanation:

The magnetic force is given by the equation

      F = q v x B

Where the blacks indicate vectors, q is the electric charge, v the velocity of the particle and B the magnetic field. The scalar magnitude of this force is

     F = q v B sin θ

Where θ is the angle between the velocity and the magnetic field.

As the force is perpendicular to the velocity, the particle describes a circular motion, using Newton's second law we can find the acceleration that is centripetal.

    F = ma

    a = v² / r

   

   q v B = mv² / r

  R = mv / qB

Let's calculate for the new speed (v₂ = 3v)

  R₂ = (m / qB) 3v

 R₂ = 3 R

 R₂ / R = 3

The new radius is three times the initial radius

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Displacement of Mr. Llama: Option D. 0 miles.

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The magnitude of the displacement of an object is equal to the distance between its final position and its initial position. In other words, as long as the initial and final positions of the object stay unchanged, the path that this object took will not affect its displacement.

For Mr. Llama:

  • Final position: Mr. Llama's house;
  • Initial position: Mr. Llama's house.

The distance between the final and initial position of Mr. Llama is equal to zero. As a result, the magnitude of Mr. Llama's displacement in the entire process will also be equal to zero.

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Two small spheres spaced 20.0 cm apart have equal charge. How many excess electrons must be present on each sphere if the magnit
Snowcat [4.5K]

Answer:

894 electrons

Explanation:

The electrostatic force between the two charges is given by:

F=\frac{k q_1 q_2}{r^2}

where we have

F=4.57\cdot 10^{-21} N is the force

k is the Coulomb's constant

q1 = q2 =q is the magnitude of the charge on each sphere

r = 20.0 cm = 0.20 m is the distance between the two spheres

Substituting and solving for q, we find the charge on each sphere:

q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(4.57\cdot 10^{-21} N)(0.20 m)^2}{9\cdot 10^9 Nm^2C^{-2}}}=1.43\cdot 10^{-16} C

And since each electron has a charge of

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the net charge on each sphere will be given by

q=Ne

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