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20 July, 20:58

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

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Answers (2)
  1. 20 July, 21:24
    0
    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
  2. 20 July, 21:51
    0
    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
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