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laila [671]
3 years ago
10

A potentialiterence of 12 volts is induced across a straight wire 0.20 meters long as it is moved at a constant speed of 3.0 met

ers per second perpendicular to a uniform magnetic field. What is the strength of the magnetic field?
A 7.27 T
B 137 T
c 20. T
D 180 T
Physics
1 answer:
shtirl [24]3 years ago
6 0

Strength of the magnetic field: 20 T

Explanation:

For a conductive wire moving perpendicular to a magnetic field, the electromotive force (voltage) induced in the wire due to electromagnetic induction is given by

\epsilon=BvL

where

B is the strength of the magnetic field

v is the speed of the wire

L is the length of the wire

For the wire in this problem, we have:

\epsilon=12 V (induced emf)

L = 0.20 m (length of the wire)

v = 3.0 m/s (speed)

Solving  for B, we find the strength of the magnetic field:

B=\frac{\epsilon}{vL}=\frac{12}{(0.20)(3.0)}=20 T

Learn more about magnetic fields:

brainly.com/question/3874443

brainly.com/question/4240735

#LearnwithBrainly

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mote1985 [20]

Answer:

The change in kinetic energy (KE) of the car is more in the second case.

Explanation:

Let the mass of the car = m

initial velocity of the first case, u = 22 km/h = 6.11 m/s

final velocity of the first case, v = 32 km/h = 8.89 m/s

change in kinetic energy (K.E) = ¹/₂m(v² - u²)

                                          ΔK.E = ¹/₂m(8.89² - 6.11²)

                                                    = 20.85m J

 

initial velocity of the second case, u = 32 km/h = 8.89 m/s

final velocity of the second case, v = 42 km/h = 11.67 m/s

change in kinetic energy (K.E) = ¹/₂m(v² - u²)

                                          ΔK.E = ¹/₂m(11.67² - 8.89²)

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The change in kinetic energy (KE) of the car is more in the second case.

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3 years ago
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