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maria [59]
3 years ago
6

A charged particle moves in a circular path in a uniform magnetic field.Which of the following would increase the period of the

particle's motion?Check all that apply.Increasing its chargeIncreasing its massIncreasing the field strengthIncreasing its speed
Physics
1 answer:
Bond [772]3 years ago
4 0

Answer:

Increasing its charge

Increasing the field strength

Explanation:

For a charged particle moving in a circular path in a uniform magnetic field, the centripetal force is provided by the magnetic force, so we can write:

qvB = m\frac{v^2}{r}

where

q is the charge

v is the velocity

B is the magnetic field

m is the mass

r is the radius of the orbit

The period of the motion is

T=\frac{2\pi r}{v}

Re-arranging for r

r=\frac{Tv}{2\pi}

And substituting into the previous equation

qvB = m \frac{Tv^3}{2\pi}

Solving for T,

T=\frac{2\pi q B}{m v^2}

So we see that the period is:

- proportional to the charge and the magnetic field

- inversely proportional to the mass and the square of the speed

So the following will increase the period of the particle's motion:

Increasing its charge

Increasing the field strength

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Dmitry [639]

Answer:

10.15Ω

Explanation:

From ohm's law,

V = IR...................... Equation 1

Where V = Voltage, I = current, R = resistance.

Assume the voltage across the resistance = V,

Given: I = 6.3 A

Substitute into equation 1

V = 6.3R.................. Equation 2

When an additional resistance of 3.4 Ω is inserted in series with R,

The voltage remain the same, but the current changes

Total Resistance(Rt) = (R+3.4)Ω, I' = 4.72 A

Also from ohm' law,

V = I'Rt............... Equation 3

Substitute the value of I'  and Rt into equation 3

V = 4.72(R+3.4)............... Equation 5.

Divide equation 2 by equation 5

V/V = 6.3R/4.72(R+3.4)

1 = 1.335R/(R+3.4)

1 = 1.335R/(R+3.4)

R+3.4 = 1.335R

3.4 = 1.335R-R

3.4 = 0.335R

R = 3.4/0.335

R = 10.15Ω

6 0
3 years ago
Read 2 more answers
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Free_Kalibri [48]

Answer: Current, resistance and voltage are the quantities which are related by Ohm's law.

Explanation:

A law which states that electric current is directly proportional to voltage and inversely proportional to resistance is called Ohm's law.

Mathematically, it is represented as follows.

I = \frac{V}{R}

where,

I = current

V = voltage

R = resistance

This means that the quantities related by Ohm's law include current, voltage and resistance.

Thus, we can conclude that current, resistance and voltage are the quantities which are related by Ohm's law.

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Explanation:

Check out the picture I drew for a minute before reading this...

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A. The distance that the racecar traveled is indeed 500m. But at the end of the lap, it is right back where it started. So overall, it has been displaced 0m.

3 0
2 years ago
A car drives 215 km east and then 45 km north. What is the magnitude of the car’s displacement? Round your answer to the nearest
34kurt

Displacement = (straight-line distance between the start point and end point) .

Since the road east is perpendicular to the road north,
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