1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
lara31 [8.8K]
3 years ago
9

An electron moves in a circular path perpendicular to a magnetic field of magnitude 0.275 T. If the kinetic energy of the electr

on is 4.90 10-19 J, find the speed of the electron and the radius of the circular path.
Physics
2 answers:
lawyer [7]3 years ago
8 0

Answer:

The speed of the electron and the radius of the circular path are 10.38\times10^{5}\ m/s and 2.15\times10^{-5}\ m.

Explanation:

Given that,

Magnetic field = 0.275 T

Kinetic energy K.E= 4.90\times10^{-19}\ J

We need to calculate the speed of the electron

Using kinetic energy

K.E=\dfrac{1}{2}mv^2

v=\sqrt{\dfrac{2K.E}{m}}

Where, m = mass of electron

K.E = kinetic energy

Put the value into the formula

v= \sqrt{\dfrac{2\times4.90\times10^{-19}}{9.1\times10^{-31}}}

v=10.38\times10^{5}\ m/s

We need to calculate the radius of the circular path

Using equation of force on charge in magnetic field

F = qvB\sin\theta....(I)

Using centripetal force

F=\dfrac{mv^2}{r}....(II)

From equation (I) and (II)

\dfrac{mv^2}{r}=qvB\sin\theta

r=\dfrac{mv^2}{qvB\sin\theta}

Put the value into the formula

r=\dfrac{9.1\times10^{-31}\times10.38\times10^{5}}{1.6\times10^{-19}\times0.275\sin90^{\circ}}

r=\dfrac{9.1\times10^{-31}\times10.38\times10^{5}}{1.6\times10^{-19}\times0.275}

r=2.15\times10^{-5}\ m

Hence, The speed of the electron and the radius of the circular path are 10.38\times10^{5}\ m/s and 2.15\times10^{-5}\ m.

xxMikexx [17]3 years ago
3 0

Answer:

Radius, r=2.14\times 10^{-5}\ m

Explanation:

It is given that,

Magnetic field, B = 0.275 T

Kinetic energy of the electron, E=4.9\times 10^{-19}\ J

Kinetic energy is given by :

E=\dfrac{1}{2}mv^2

v=\sqrt{\dfrac{2E}{m}}

v=\sqrt{\dfrac{2\times 4.9\times 10^{-19}}{9.1\times 10^{-31}}}            

v = 1037749.04 m/s

The centripetal force is balanced by the magnetic force as :

qvB\ sin90=\dfrac{mv^2}{r}

r=\dfrac{mv}{qB}

r=\dfrac{9.1\times 10^{-31}\times 1037749.04}{1.6\times 10^{-19}\times 0.275 }

r=2.14\times 10^{-5}\ m

So, the radius of the circular path is 2.14\times 10^{-5}\ m. Hence, this is the required solution.

You might be interested in
How do you find circular motion?​
MatroZZZ [7]

Answer:

Explanation:

By multiplying the rotational frequency with the circumference we can determine the average speed of the object. The circular velocity formula is expressed as, vc = 2 πr / T. Where in, r denotes the radius of the circular orbit. T is time period.

6 0
3 years ago
Read 2 more answers
Two resistors, R1=3.85 Ω and R2=6.47 Ω , are connected in series to a battery with an EMF of 24.0 V and negligible internal resi
Bond [772]

Answer:

(a) 2.33 A

(b) 15.075 V

Explanation:

From the question,

The total resistance (Rt) = R1+R2 = 3.85+6.47

R(t) = 10.32 ohms.

Applying ohm's law,

V = IR(t)..........equation 1

Where V = Emf of the battery, I = current flowing through the circuit, R(t) = combined resistance of both resistors.

Note: Since both resistors are connected in series, the current flowing through them is the same.

Therefore,

I = V/R(t)............. Equation 2

Given: V = 24 V, R(t) = 10.32 ohms

Substitute these values into equation 2

I = 24/10.32

I = 2.33 A.

Hence the current through R1 = 2.33 A.

V2 = IR2.............. Equation 3

V2 = 2.33(6.47)

V2 = 15.075 V

7 0
3 years ago
The period of time required for the moon to complete a cycle of phases is called the ________ month.
eduard
Synodic month, also known as a lunar month.
3 0
3 years ago
The frequency of a wave is 560 Hz. What is it’s period
Crazy boy [7]
The answer would be, "1/560 seconds".
4 0
3 years ago
A raft with the area A , thickness= h and the mass 600 kg, Floats in still water with 7 cm
elena55 [62]

<span>In this problem, we need to solve for Bubba’s mass. To do this, we let A be the area of the raft and set the weight of the displaced fluid with the raft alone as ρwAd1g and ρwAd2g with the person on the raft, </span>where ρw is the density of water, d1 = 7cm, and d2= 8.4 cm. Set the weight of displaced fluid equal to the weight of the floating objects to eliminate A and ρw then solve for m.

<span>ρwAd1g = Mg</span>

ρw<span>Ad2g = (M + m) g</span>

<span>d2∕d1 = (M + m)/g</span>

m = [(d2<span>∕d1)-1] M = [(8.4 cm/7.0 cm) - 1] (600 kg) =120 kg</span>

This means that Bubba’s mass is 120 kg.

7 0
3 years ago
Other questions:
  • Which characteristics do all terrestrial planets have?
    13·2 answers
  • Is diet Pepsi heterogeneous or is it homogeneous
    8·2 answers
  • Which is the SI base unit for distance?
    5·1 answer
  • A tennis ball pitching machine goes haywire and pitches at 10 rounds per second and the speed is an incredible 300m/s, what is t
    7·2 answers
  • Which will increase the energy of motion of water molecules?
    12·1 answer
  • The Pioneer 10 spacecraft was launched in March 1972 to explore the solar system. Pioneer 10 has continued on its journey and is
    8·2 answers
  • A scientist that applies the laws of science to the needs of communities is called _____.
    9·1 answer
  • A 600kg car is at rest, and then it accelerates to 5 m/s.
    5·1 answer
  • A small first-aid kit is dropped by a rock climber who is descending steadily at -1.25 m/s. After 2.5 seconds, what is the veloc
    14·1 answer
  • A classic demonstration illustrating eddy currents is performed by dropping a permanent magnet inside a conducting cylinder. The
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!