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chubhunter [2.5K]
3 years ago
5

what magnetic field is needed to exert a force of 0.4N on a 0.1m long conductor if it carries a current of 2A perpendicular to t

he magnetic field?
Physics
1 answer:
serious [3.7K]3 years ago
6 0

Answer:

2T

Explanation:

Use F = BIL

F = 0.4N

B = ??

I = 2A

L = 0.1m

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An electron is initially at rest in a uniform electric field having a strength of 1.85 × 106 V/m. It is then released and accele
kirza4 [7]

Answer:

W = 462.5 keV

Explanation:

As we know that when electron moved in electric field then work done by electric field must be equal to the change in kinetic energy of the electron

So here we have to find the work done by electric field on moving electron

So we have

F = qE

F = (1.6 \times 10^{-19})(1.85 \times 10^6)

F = 2.96 \times 10^{-13} N

now the distance moved by the electron is given as

d = 0.25 m

so we have

W = F.d

W = (1.6 \times 10^{-19})(1.85 \times 10^6)(0.25)

W = 7.4 \times 10^{-14} J

now we have to convert it into keV units

so we have

1 keV = 1.6 \times 10^{-16} J

W = 462.5 keV

5 0
3 years ago
Consider your knowledge and experience about caring for children. What have you observed about the needs, behaviors, and abiliti
Advocard [28]
They all have a special way in their age how they want to get taught things but there abilities make them special
5 0
3 years ago
During a neighborhood baseball game in a vacant lot, a particularly wild hit sends a 0.146 kg baseball crashing through the pane
Nonamiya [84]

Answer:

Impulse, |J| = 0.6716 kg-m/s

Force, F = 63.35 N

Explanation:

It is given that,

Mass of the baseball, m = 0.146 kg

Initial speed of the ball, u = 15.3 m/s

Final speed of the ball, v = 10.7 m/s

To find,

(a) The magnitude of this impulse.

(b) The magnitude of the average force of the glass on the ball.

Solution,

(a) Impulse of an object is equal to the change in its momentum. It is given by :

J=m(v-u)

J=0.146\ kg(10.7-15.3)\ m/s

J = -0.6716 kg-m/s

or

|J| = 0.6716 kg-m/s

(b) Another definition of impulse is given by the product of force and time of contact.

t = 0.0106 s

J=F\times \Delta t

F=\dfrac{J}{\Delta t}

F=\dfrac{0.6716\ kg-m/s}{0.0106\ s}

F = 63.35 N

Hence, this is the required solution.

6 0
3 years ago
Your friend decides to generate electrical power by rotating a 100,000 turn coil of wire around an axis in the plane of the coil
MakcuM [25]

Answer:

a) I=35mA

b) P=1.73W

Explanation:

a) The max emf obtained in a rotating coil of N turns is given by:

emf_{max}=NBA\omega

where N is the number of turns in the coil, B is the magnitude of the magnetic field, A is the area and w is the angular velocity of the coil.

By calculating A and replacing in the formula (1G=10^{-4}T) we get:

A=\pi r^2 =\pi(0.23m)^2=0.16m^2

emf_{max}=(100000)(0.3*10^{-4}T)(0.166m^2)(140\frac{rev}{s})=69.72V

Finally, the peak current is given by:

I=\frac{emf}{R}=\frac{69.72V}{1400\Omega}=49.8mA

b)

we have that

I_{rms}=\frac{I}{\sqrt{2}}=\frac{0.0498A}{\sqrt{2}}=0.035A

P_{rms}=I^2{rms}R=(0.035A)^2(1400\Omega)=1.73W

hope this helps!!

6 0
3 years ago
Pls help I’m in a middle of a test and I can’t find the answer
EastWind [94]

Answer:

I believe the answer is B.

5 0
1 year ago
Read 2 more answers
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