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

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,

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

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

⇒      $=2240\times \pi \ 0.000001$

⇒      $=7.037\times 10^{-3} \ kg$