Question

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

Answer 1

Answer:

5.099 ohm

Explanation:

Let number of the turn of wire be

N = L/D

Where L is the length of the solenoid and d is the diameter 75 cm = 0.75m , 0.100 = 0.001m

N = 0.75/0.100

N = 750

Hence the number of turns of copper wire is 750

The expression for the length of the wire

L = N ( πD )

Where D is the diameter of the solenoid. And take 750 for N and 10.0cm for the diameter.

Hence, L = 750 ( π × 10/100 )

L = 750 ×3.142× 0.1

L = 235.65m

The resistance of the wire is given as : R = p 4l/πd×d

Where p is the resistivity of the wire

l is the length of the wire and

d is the diameter of the wire

Let l = 235.65m , p = 1.7×10~ -8

0.100cm = 0.001m for the diameter

R = 1.7×10~-8 { 4× 235.65 } / π × ( 0.001m* 0.001 m)

R = 0.000016024/ 3.142 × 0.000001

R = 0.000016024/0.000003142

R = 5.099 ohms

Answer 2

Answer:

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