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adelina 88 [10]
3 years ago
15

An electromagnet produces a magnetic eld of 0.95 T in a cylindrical region of radius 3.0 cm between its poles. A straight wire c

arrying a current of 15.0 A passes through the center of this region and is perpendicular to both the axis of the cylindrical region and the magnetic eld. What magnitude of force is exerted on the wire?
Physics
1 answer:
Daniel [21]3 years ago
6 0

Answer:

0.855 N

Explanation:

Using

F = BILsinФ.................... Equation 1

Where F = magnitude of the force exerted on the wire, B = magnetic Field, I = current, L = Length of the wire in the magnetic Field, Ф = angle between the wire and the magnetic field.

Given: B = 0.95 T, I = 15.0 A, L = 2r, r = radius = 3.0 cm = 0.03 m L = 2×0.03 = 0.06 m, Ф = 90° (perpendicular)

Substitute into equation 1

F = 0.95(15)(0.06)sin90°

F = 0.855 N.

Hence the magnitude of force that is exerted on the wire = 0.855 N

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What is the potential energy of the system composed of the three charges q1, q3, and q4, when q1 is at point R? Define the poten
Ad libitum [116K]

Answer:

Incomplete question check attachment for complete question

Also it is given that

q1=-0.7uC

q4=-1.7uC

q3=-1.7uC

Also the distance are given as

a=2.2cm=0.022m

d2=3.6cm=0.036m

Explanation:

The potential energy due to point R is given as

The potential energy due to charge q1 and q3 plus the potential energy due to charge q4 and q1 plus the potential energy due to charge q3 and q4

So, let take it one after the other

Potential energy is give as

P.E=kq1q2/r

Therefore,

Potential energy due to charge q1 and q3

U¹³=kq1q3/r

To get the distance between charge q1 and q3, we will apply Pythagoras theorem

r=√(d2²+a²)

r=√(0.036²+0.022²)

r=0.0422m

k is a constant =9×10^9Nm²/C²

Then,

U¹³=kq1q3/r

U¹³=9×10^9×0.7×10^-6×1.7×10^-6/0.0422

U¹³=0.254J

Potential energy due to charge q1 and q4

U¹⁴=kq1q4/r

To get the distance between charge q1 and q4, we will apply Pythagoras theorem

r=√(d2²+a²)

r=√(0.036²+0.022²)

r=0.0422m

k is a constant =9×10^9Nm²/C²

Then,

U¹⁴=kq1q4/r

U¹⁴=9×10^9×0.7×10^-6×1.7×10^-6/0.0422

U¹⁴=0.254J

Potential energy due to charge q3 and q4

U³⁴=kq3q4/r

r=2a=2×0.022=0.044m

k is a constant =9×10^9Nm²/C²

Then,

U³⁴=kq3q4/r

U³⁴=9×10^9×1.7×10^-6×1.7×10^-6/0.044

U¹⁴=0.591J

Then, the total energy is

U= U¹³+ U¹⁴ + U³⁴

U=0.254+0.254+0.591

U=1.099J

Then also, the potential energy is zero because at infinity both U¹³ and U¹⁴ will have infinite potential because their distance apart will be infinite.

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3 years ago
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Answer:

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Explanation:

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3 years ago
A car's acceleration is 3m/s2. If the car started at rest and it only took 10s for the car to reach this acceleration, what is t
Grace [21]

Answer:

30m/s

Explanation:

From law of motion equation

Vf= Vi + at

Where Vf= final velocity

Vi= initial velocity=0(the car started at rest)

a= acceleration= 3m/s2

t= time= 10s

Then substitute into the equation to get the final velocity.

Vf= 0+(10×3)

Vf= 30m/s

Hence, the car's final velocity is 30m/s

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3 years ago
How can we calculate the e.m.f of the battery?.
cluponka [151]

Explanation:

The emf is equal to the work done on the charge per unit charge (ϵ=dWdq) when there is no current flowing. Since the unit for work is the joule and the unit for charge is the coulomb, the unit for emf is the volt (1V=1J/C).

7 0
3 years ago
Read 2 more answers
A student walks 50 meters east, 40 meters north, 35 meters east, and then 20 m south. What is the magnitude and direction of the
Nuetrik [128]

Answer: A student walks 50 meters east, 40 meters north, 35 meters east, and then 20 m south. Then the magnitude and direction of the student's total displacement will be 87.32 m along the direction of AD or in east-south direction.

Explanation: To find the correct answer, we need to know about the Displacement of a body in motion.

<h3>What is displacement of a body in motion?</h3>
  • The displacement is the shortest distance between initial and final positions of a body.
  • It's a vector quantity, and can positive, negative, or zero.
  • The magnitude of displacement is less than or equal to the distance travelled.
<h3>How to solve the problem?</h3>
  • At first, we can draw a diagram showing the motion of the body.
  • From the diagram, the displacement of the body will be equal to the distance between point A and D.
  • To solve this, we can use Pythagoras theorem.

AD=AC+CD\\AC^{2} =50^{2} +11^2\\AC=51.19 m\\Similarly,\\CD^2=35^2+9^2\\CD=36.13 m\\thus, \\AD=51.19+36.13=87.32 m

Thus, from the above calculations, we can conclude that, the displacement of the body will be equal to 87.32 m along the direction of AD or in east-south direction.

Learn more about the Displacement here:

brainly.com/question/28020108

#SPJ4

3 0
2 years ago
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