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zloy xaker [14]
3 years ago
13

Examples of field force

Physics
1 answer:
Sindrei [870]3 years ago
7 0
Here are some examples of field force:
1. <span>1) electrostatic force of attraction or repulsion on a charged particle in an electric field
2. </span><span>) magnetic force on charged particles moving at right angles with a magnetic field. Magnetic force also affects magnetic poles.


I hope this helps. </span>
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Physics B 2020 Unit 3 Test
weqwewe [10]

Answer:

1)

When a charge is in motion in a magnetic field, the charge experiences a force of magnitude

F=qvB sin \theta

where here:

For the proton in this problem:

q=1.602\cdot 10^{-19}C is the charge of the proton

v = 300 m/s is the speed of the proton

B = 19 T is the magnetic field

\theta=65^{\circ} is the angle between the directions of v and B

So the force is

F=(1.602\cdot 10^{-19})(300)(19)(sin 65^{\circ})=8.28\cdot 10^{-16} N

2)

The magnetic field produced by a bar magnet has field lines going from the North pole towards the South Pole.

The density of the field lines at any point tells how strong is the magnetic field at that point.

If we observe the field lines around a magnet, we observe that:

- The density of field lines is higher near the Poles

- The density of field lines is lower far from the Poles

Therefore, this means that the magnetic field of a magnet is stronger near the North and South Pole.

3)

The right hand rule gives the direction of the  force experienced by a charged particle moving in a magnetic field.

It can be applied as follows:

- Direction of index finger = direction of motion of the charge

- Direction of middle finger = direction of magnetic field

- Direction of thumb = direction of the force (for a negative charge, the direction must be reversed)

In this problem:

- Direction of motion = to the right (index finger)

- Direction of field = downward (middle finger)

- Direction of force = into the screen (thumb)

4)

The radius of a particle moving in a magnetic field is given by:

r=\frac{mv}{qB}

where here we have:

m=6.64\cdot 10^{-22} kg is the mass of the alpha particle

v=2155 m/s is the speed of the alpha particle

q=2\cdot 1.602\cdot 10^{-19}=3.204\cdot 10^{-19}C is the charge of the alpha particle

B = 12.2 T is the strength of the magnetic field

Substituting, we find:

r=\frac{(6.64\cdot 10^{-22})(2155)}{(3.204\cdot 10^{-19})(12.2)}=0.366 m

5)

The cyclotron frequency of a charged particle in circular motion in a magnetic field is:

f=\frac{qB}{2\pi m}

where here:

q=1.602\cdot 10^{-19}C is the charge of the electron

B = 0.0045 T is the strength of the magnetic field

m=9.31\cdot 10^{-31} kg is the mass of the electron

Substituting, we find:

f=\frac{(1.602\cdot 10^{-19})(0.0045)}{2\pi (9.31\cdot 10^{-31})}=1.23\cdot 10^8 Hz

6)

When a charged particle moves in a magnetic field, its path has a helical shape, because it is the composition of two motions:

1- A uniform motion in a certain direction

2- A circular motion in the direction perpendicular to the magnetic field

The second motion is due to the presence of the magnetic force. However, we know that the direction of the magnetic force depends on the sign of the charge: when the sign of the charge is changed, the direction of the force is reversed.

Therefore in this case, when the particle gains the opposite charge, the circular motion 2) changes sign, so the path will remains helical, but it reverses direction.

7)

The electromotive force induced in a conducting loop due to electromagnetic induction is given by Faraday-Newmann-Lenz:

\epsilon=-\frac{N\Delta \Phi}{\Delta t}

where

N is the number of turns in the loop

\Delta \Phi is the change in magnetic flux through the loop

\Delta t is the time elapsed

From the formula, we see that the emf is induced in the loop (and so, a current is also induced) only if \Delta \Phi \neq 0, which means only if there is a change in magnetic flux through the loop: this occurs if the magnetic field is changing, or if the area of the loop is changing, or if the angle between the loop and the field is changing.

8)

The flux is calculated as

\Phi = BA sin \theta

where

B = 5.5 T is the strength of the magnetic field

A is the area of the coil

\theta=18^{\circ} is the angle between the  direction of the field and the plane of the loop

Here the loop is rectangular with lenght 15 cm and width 8 cm, so the area is

A=(0.15 m)(0.08 m)=0.012 m^2

So the flux is

\Phi = (5.5)(0.012)(sin 18^{\circ})=0.021 Wb

See the last 7 answers in the attached document.

Download docx
<span class="sg-text sg-text--link sg-text--bold sg-text--link-disabled sg-text--blue-dark"> docx </span>
<span class="sg-text sg-text--link sg-text--bold sg-text--link-disabled sg-text--blue-dark"> pdf </span>
5 0
3 years ago
A light-year is about?
denis-greek [22]
13 year old and the light year mean  cell of moucles that can use simple year light diffusion
7 0
3 years ago
Your laundry basket weighs 22 N and your room is 3.0 m above you on the second floor. It takes you 6.0 seconds to carry the laun
Fofino [41]
<h2>Power is 11 W</h2>

Explanation:

Power = Work ÷ Time

Work = Force x Displacement

Force = 22 N

Displacement = 3 m

Time = 6 seconds

Substituting

       Work = Force x Displacement

       Work = 22 x 3 = 66 J

       Power = Work ÷ Time

       Power = 66 ÷ 6

       Power = 11 W

Power is 11 W

8 0
3 years ago
1) A force of 20 Newton acts on a bar having a cross sectional area of 0.8m^2 and length 10cm.calculate the stress developed in
Elanso [62]

Answer:25N/M^2

Explanation:

Force=20N Area=0.8M^2

Stress=force/area

Stress=20/0.8

Stress=25N/M^2

4 0
3 years ago
The gravitational attraction between two objects will what is the object move further apart
finlep [7]
The farther apart the two objects, the weaker the gravitational attraction between them.
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