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WINSTONCH [101]
2 years ago
14

A 0.43 m long and 0.43 m wide loop is moved at a constant velocity of 0.15 m/s into a perpendicular constant magnetic field of 0

.31 T. Calculate the magnitude of the induced voltage in the loop.
Physics
1 answer:
olya-2409 [2.1K]2 years ago
7 0

Answer:

The magnitude of the induced voltage in the loop is 20 mV.

Explanation:

given;

length of loop, L = 0.43 m

width of loop,w = 0.43 m

velocity of moved loop, v = 0.15m/s

magnetic field strength,B = 0.31 T

To determine the magnitude of the induced voltage in the loop, we apply Faraday's law;

magnitude induced E.M.F = BLv

magnitude induced E.M.F = 0.31 x 0.43 x 0.15 = 0.02 V = 20 mV

Therefore, the magnitude of the induced voltage in the loop is 20 mV.

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Explanation:

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A long uniformly charged thread (linear charge density λ= 2.5 C/m) lies along the x axis in the figure.(Figure 1) A small charge
Kamila [148]

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Answer 2) E_{q}  = {2.3 / (4 2 π ε°) ( - 7/ (√(84); -12 / (√84))

Adding these two vectors will give the length which is magnitude of the combined field.  

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Please refer the graph and the attachment for better understanding.

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A 1.00-kg mass is attatched to a string 1.0m long and completes a horizontal circle in 0.25s. What is the centripetal accelerati
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</span>
That's it.  It takes roughly a 142-pound pull on the string to keep
1 kilogram revolving at a 1-meter radius 4 times a second !<span> 
</span>
If you eased up on the string, the kilogram could keep revolving
in the same circle, but not as fast.

You also need to be very careful with this experiment, and use a string
that can hold up to a couple hundred pounds of tension without snapping. 
If you've got that thing spinning at 4 times per second and the string breaks,
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be hazardous to the health of anybody who's been watching you and wondering
what you're doing.


3 0
3 years ago
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