Question

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 field 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?

Answer 1

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

Step-by-step 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* I* L=M* g`

On putting the estimated values, we get

⇒
`B* 50* 1=7.037* 10^(-3)* 9.81`

⇒
`50B=69.03297* 10^(-3)`

⇒
`B=1.38* 10^(-3) \ T`

(b)...

As we know,

⇒
`m=\delta* L* (\pi \ d^2)/(4)`

⇒
`=8960* 1* (\pi \ (0.001)^2)/(4)`

⇒
`=2240* \pi \ 0.000001`

⇒
`=7.037* 10^(-3) \ kg`