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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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Reference the diagram below for clarification.

1.

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In parallel:
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\frac{1}{R_{T, parallel}} = \frac{1}{10} + \frac{1}{10}\\\\\frac{1}{R_{T, parallel}} = \frac{2}{10} = \frac{1}{5}\\\\R_{T, parallel} = 5\Omega

Now that we have reduced the parallel resistors to a 'single' resistor, we can add their equivalent resistance with the other resistor in parallel (15 Ohm) using series rules:
R_T = 15 + 5\\\\\boxed{R_T = 20 \Omega}

2.

We can use Ohm's law to solve for the current in the circuit.

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3.

For resistors in series, both resistors receive the SAME current.

Therefore, the 15Ω resistor receives 6A, and the parallel COMBO (not each individual resistor, but the 5Ω equivalent when combined) receives 6A.

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Since the 15.0 Ω resistor receives 6A, we can use Ohm's Law to solve for voltage.

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Part b)

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