Question

As part of an advanced lab project, you design and construct a solenoid by wrapping a single layer of copper wire with a thin layer of varnish as insulation and with a diameter of 0.120 cm onto a plastic tube. The tube has a diameter of 12.0 cm, a length of 85.5 cm, and the coils are wrapped as tightly as possible. Determine the power that must be delivered to the solenoid in order to create a magnetic field of 9.60 mT at its center. The resistivity for copper is

Answer 1

Answer:

P = 0.3668 W = 366.8 mW

Explanation:

First we calculate the total no. of coils of solenoid:

N = L/d

where,

N = Total No. of Coils = ?

L = Length of Tube = 85.5 cm = 0.855 m

d = diameter of wire = 0.12 cm = 0.0012 m

Therefore,

N = 85.5 cm/0.12 cm

N = 712.5 = 712 turns

Now, we use formula for magnetic field strength of solenoid:

B = μ₀NI/L

where,

B = Magnetic Field = 9.6 mT = 0.0096 T

μ₀ = 4π x 10⁻⁷ N/A²

I = Current = ?

Therefore,

0.0096 T = (712)(4π x 10⁻⁷ N/A²)(I)/(0.855 m)

I = 0.0096/0.00104

I = 9.17 A

Now we will find resistance of wire:

R = ρLw/A

where,

R = Resistance of copper wire = ?

ρ = resistivity of copper = 1.7 x 10⁻⁸ Ωm

Lw = Length of wire used = 2πrN = πdN = π(0.0012 m)(712) = 2.68 m

A=cross-sectional area of wire = πr² = πd²/4 = π(0.0012 m)²/4 = 1.13 x 10⁻⁶ m²

Therefore,

R = (1.7 x 10⁻⁸ Ωm)(2.68 m)/(1.13 x 10⁻⁶ m²)

R = 0.04 Ω

No, the delivered can be calculated as:

P = VI = (IR)I = I²R

P = (9.17 A)²(0.04 Ω)

P = 0.3668 W = 366.8 mW