A solenoid 10.0 cm in diameter and 75.0 cm long is made from copper wire of diameter 0.100 cm, with very thin insulation. The wire is wound onto a cardboard tube in a single layer, with adja- cent turns touching each other. What power must be delivered to the solenoid if it is to produce a field of 8.00 mT at its center

Power = 2.22W

Explanation:

We are given;

Length of Solenoid; L = 75cm = 0.75m

Diameter of copper wire; d = 0.1 cm = 0.001m

B = 8 mT = 0.008T

From Ampere's law formula,

B = μo•I•N/L

Where;

B is magnetic field

I is current

N is number of turns

L is length of Solenoid

μo is vacuum permeability amd it has a constant value of 4π x 10^(-7) H/m

Now, let's find N.

N is given by; N = L/d

N = 0.75/0.001 = 750 turns

From earlier, B = μo•I•N/L

Thus, let's make I the subject;

I = B•L/(μo•N)

I = (0.008 x 0.75)/(4π x 10^(-7) x 750)

I = 0.006/(0.0009424778) = 6.367 A

Now, let's find the resistance; R.

The resistance; R is given by the formula;

R = ρ•N•π•d/A

Where;

ρ is resistivity of copper wire and has a value of 1.68 x 10^(-8) Ω.m

A is the area

Now area is given as; A = πd²/4

Putting this in the resistance equation gives;

R = ρ•N•π•d/(πd²/4)

This gives;

R = (4ρ•N•π•d)/(πd²)

This leads to; R = (4ρ•N)/(d)

Plugging in the relevant values, we have;

R = (4 x 1.68 x 10^(-8) x 750)/0.001

R = 0.0504 Ω

Now, we know that the formula for Power is;

P = I²R

where I is current and R is resistance.

Thus, Power = 6.637² x 0.0504 = 2.22 W

   F = 93.25 lb

Explanation:

To solve this problem let's use the rotational equilibrium condition

         Σ τ = 0

let's fix a reference system located at the end of the table, the position of Juan and assume the counterclockwise turns as positive

        F_jhon 0 - w l / 2 - W (l-4.9) + F l = 0

        F = w l / 2 + W (l -4.9) / l

where w is the weight of the table and W the weight of the box

let's calculate

        F = (44.5 12/2 + 120 (12- 4.9)) / 12

        F = 93.25 lb