Ask question

Ask question

All categories

4 years ago

12

What is the speed of a high-speed electron traveling at right angles to a 0.80-tesla magnetic field with a force of -7.5 x 10^-1

2 newtons? The charge of an electron is 1.60 x 10^-19

1 answer:

serg [7]4 years ago

7 0

Explanation:

Answer:

Magnetic field, B = 0.016 Tesla

Explanation:

It is given that,

Velocity of electron, v=9.6\times 10^5\ m/sv=9.6×10

5

m/s

Magnetic force, F=2.6\times 10^{-15}\ NF=2.6×10

−15

N

Charge, q=1.6\times 10^{-19}\ Nq=1.6×10

−19

N

The magnetic force is given by :

F=qvBF=qvB

B=\dfrac{F}{qv}B=

qv

F

B=\dfrac{2.6\times 10^{-15}}{1.6\times 10^{-19}\times 9.6\times 10^5}B=

1.6×10

−19

×9.6×10

5

2.6×10

−15

B=0.016\ TB=0.016 T

So, the magnitude of magnetic field is 0.016 Tesla. Hence, this is the required solution.

You might be interested in

A sound wave produced by a loudspeaker can travel through water, but not through a vacuum. In comparison, a red light wave produ

Anna [14]

Answer:

A. Light can travel trough air and space.

Explanation:

A vacuum is space that has no air inside of it. Since light is able to go through 'nothing' and matter, your answer must be A.

8 0

3 years ago

A heating element having a resistance (at its operating temperature) of 148 Ω is connected to a battery having an emf of 523 V a

vladimir2022 [97]

Answer:

The correct answer is 283 watts

4 0

4 years ago

Read 2 more answers

A coaxial cable consists of alternating coaxial cylinders of conducting and insulating material.

olga2289 [7]

Answer:

True

Explanation:

A coaxial cable is a type of cable that has an inner conductor surrounded by an insulating layer, surrounded by a conductive shielding.

4 0

3 years ago

A physical therapy exercise has a person shaking a 5.00 kg weight up and down rapidly. if the barbell is moving at 4.50 m/s, wha

Ivanshal [37]

The magnitude of the force required to stop the weight in 0.333 seconds is 67.6 N.

<h3>Magnitude of required force to stop the weight</h3>

The magnitude of the force required to stop the weight in 0.333 seconds is calculated by applying Newton's second law of motion as shown below;

F = ma

F = m(v/t)

F = (mv)/t

F = (5 x 4.5)/0.333

F = 67.6 N

Thus, the magnitude of the force required to stop the weight in 0.333 seconds is 67.6 N.

Learn more about force here: brainly.com/question/12970081

#SPJ1

7 0

2 years ago

Coulomb’s law and static point charge ensembles (15 points). A test charge of 2C is located at point (3, 3, 5) in Cartesian coor

fenix001 [56]

Answer:

a) $F_{r}= -583.72MN i + 183.47MN j + 6.05GN k$

b) $E=3.04 \frac{GN}{C}$

Step-by-step explanation.

In order to solve this problem, we mus start by plotting the given points and charges. That will help us visualize the problem better and determine the direction of the forces (see attached picture).

Once we drew the points, we can start calculating the forces:

$r_{AP}^{2}=(3-0)^{2}+(3-0)^{2}+(5+0)^{2}$

which yields:

$r_{AP}^{2}= 43 m^{2}$

(I will assume the positions are in meters)

Next, we can make use of the force formula:

$F=k_{e}\frac{q_{1}q_{2}}{r^{2}}$

so we substitute the values:

$F_{AP}=(8.99x10^{9})\frac{(1C)(2C)}{43m^{2}}$

which yields:

$F_{AP}=418.14 MN$

Now we can find its components:

$F_{APx}=418.14 MN*\frac{3}{\sqrt{43}}i$

$F_{APx}=191.30 MNi$

$F_{APy}=418.14 MN*\frac{3}{\sqrt{43}}j$

$F_{APy}=191.30MN j$

$F_{APz}=418.14 MN*\frac{5}{\sqrt{43}}k$

$F_{APz}=318.83 MN k$

And we can now write them together for the first force, so we get:

$F_{AP}=(191.30i+191.30j+318.83k)MN$

We continue with the next force. The procedure is the same so we get:

$r_{BP}^{2}=(3-1)^{2}+(3-1)^{2}+(5+0)^{2}$

which yields:

$r_{BP}^{2}= 33 m^{2}$

Next, we can make use of the force formula:

$F_{BP}=(8.99x10^{9})\frac{(4C)(2C)}{33m^{2}}$

which yields:

$F_{BP}=2.18 GN$

Now we can find its components:

$F_{BPx}=2.18 GN*\frac{2}{\sqrt{33}}i$

$F_{BPx}=758.98 MNi$

$F_{BPy}=2.18 GN*\frac{2}{\sqrt{33}}j$

$F_{BPy}=758.98MN j$

$F_{BPz}=2.18 GN*\frac{5}{\sqrt{33}}k$

$F_{BPz}=1.897 GN k$

And we can now write them together for the second, so we get:

$F_{BP}=(758.98i + 758.98j + 1897k)MN$

We continue with the next force. The procedure is the same so we get:

$r_{CP}^{2}=(3-5)^{2}+(3-4)^{2}+(5-0)^{2}$

which yields:

$r_{CP}^{2}= 30 m^{2}$

Next, we can make use of the force formula:

$F_{CP}=(8.99x10^{9})\frac{(7C)(2C)}{30m^{2}}$

which yields:

$F_{CP}=4.20 GN$

Now we can find its components:

$F_{CPx}=4.20 GN*\frac{-2}{\sqrt{30}}i$

$F_{CPx}=-1.534 GNi$

$F_{CPy}=4.20 GN*\frac{2}{\sqrt{30}}j$

$F_{CPy}=-766.81 MN j$

$F_{CPz}=4.20 GN*\frac{5}{\sqrt{30}}k$

$F_{CPz}=3.83 GN k$

And we can now write them together for the third force, so we get:

$F_{CP}=(-1.534i - 0.76681j +3.83k)GN$

So in order to find the resultant force, we need to add the forces together:

$F_{r}=F_{AP}+F_{BP}+F_{CP}$

so we get:

$F_{r}=(191.30i+191.30j+318.83k)MN + (758.98i + 758.98j + 1897k)MN + (-1.534i - 0.76681j +3.83k)GN$

So when adding the problem together we get that:

$F_{r}=(-0.583.72i + 0.18347j +6.05k)GN$

which is the answer to part a), now let's take a look at part b).

b)

Basically, we need to find the magnitude of the force and divide it into the test charge, so we get:

$F_{r}=\sqrt{(-0.583.72)^{2} + (0.18347)^{2} +(6.05)^{2}}$

which yields:

$F_{r}=6.08 GN$

and now we take the formula for the electric field which is:

$E=\frac{F_{r}}{q}$

so we go ahead and substitute:

$E=\frac{6.08GN}{2C}$

$E=3.04\frac{GN}{C}$

7 0

4 years ago