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1 year ago

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A 1.0-m-long copper wire of diameter 0.10 cm carries a current of 50.0 A to the east. Suppose we apply to this wire a magnetic f

ield that produces on it an upward force exactly equal in magnitude to the wire's weight, causing the wire to "levitate."
Required:
a. What is the field's magnitude?
b. What is the field's direction?

1 answer:

Setler79 [48]1 year ago

7 0

Answer:

The classification of that same issue in question is characterized below.

Explanation:

The given values are:

Current, I = 50.0 A

Diameter, d = 0.10 cm

(a)...

As we know,

⇒ Magnetic force = Copper wire's weight

So,

⇒ $B\times I\times L=M\times g$

On putting the estimated values, we get

⇒ $B\times 50\times 1=7.037\times 10^{-3}\times 9.81$

⇒ $50B=69.03297\times 10^{-3}$

⇒ $B=1.38\times 10^{-3} \ T$

(b)...

As we know,

⇒ $m=\delta\times L\times \frac{\pi \ d^2}{4}$

⇒ $=8960\times 1\times \frac{\pi \ (0.001)^2}{4}$

⇒ $=2240\times \pi \ 0.000001$

⇒ $=7.037\times 10^{-3} \ kg$

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Answer:

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$Percent Error=\frac{V_{measured}- V_{true} } {V_{true}} *100$

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First, evaluate the fraction. Subtract 2.00 from 2.02

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Explanation:

Given:

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$x$ when the spring is compressed

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