A solenoid is created by wrapping a L = 55 m long wire around a hollow tube of diameter D = 4.5 cm. The wire diameter is d = 0.55 mm. The solenoid wire is then connected to a power supply so that a current of I = 8.5 A flows through the wire.

A. Write an expression for the number of turns, N, in the solenoid. You do not need to take into account the diameter of the wire in this calculation.

B. Calculate the number of turns, N, in the solenoid.

C. Write an expression for the length of the solenoid, (L2), in terms of the hollow tube D, the length of the wire L, and the diameter of the wire d. Assume it is constructed by using only 1 layer of loops (note that most solenoids are actually constructed using many layers to maximize magnetic field density).

D. Calculate the length of the solenoid, (L2), in meters.

E. Calculate the magnitude of the magnetic field at the center of the solenoid, in Tesla.

A. N = L/πD

B. N  = 389 turns

C. L₂ = DN

D. L₂ = 17.5 m

E. B = 2.3 x 10⁻⁴ Web

Explanation:

A.

For each number of turn a length of wire equal to the circumference of tube. Therefore, the no. of turns can be given as:

N = L/2πr

N = L/πD

where,

N = No. of turns

D = Diameter of tube

L = Length of wire

B.

using values in the equation, we get:

N = 55 m/π(0.045 m)

N = 389 turns

C.

Since, there is only one layer of loop. Therefore, the length of solenoid can be easily found out by adding the diameter of wire for each turn. Hence:

L₂ = DN

where,

L₂ = Length of the solenoid

D = diameter of wire

N = No. of Turns

D.

using values in the equation we get:

L₂ = (0.045 m)(389)

L₂ = 17.5 m

E.

The magnitude of magnetic field inside solenoid is given by the formula:

B = μ₀NI/L₂

where,

B = Magnetic field inside solenoid

μ₀ = permeability = 4π x 10⁻⁷ T/A.m

I = current = 8.5 A

Therefore,

B = (4π x 10⁻⁷ T/A.m)(389)(8.5 A)/17.5 m)

B = 2.3 x 10⁻⁴ Web

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