A Current-Carrying Coil as a Magnetic Dipole Problems and Solutions

 Problem#1

Figure 1 shows an arrangement known as a Helmholtz coil. It consists of two circular coaxial coils, each of 200 turns and radius R = 25.0 cm, separated by a distance s = R. The two coils carry equal currents i = 12.2 mA in the same direction. Find the magnitude of the net magnetic field at P, midway between the coils.

Fig.1

Answer:
Let us apply the law of Biot and Savart to a differential element ds of the loop

dB = (μ₀I cos α)ds/4πr²

Figure 2 shows that r and α are related to each other. Let us express each in terms of the variable x, the distance between point P and the center of the loop. The relations are

Fig.2

r = (R² + x²)^1/2

and
cos α = R/r = R/(R² + x²)^1/2

So that
B = {μ₀IR/[4π(R² + x²)^3/2]} ∫ds

or, because is simply the circumference 2πR of the loop,

B(x) = μ₀iR²/{2(R² + x²)^3/2}

The magnitude field a distance x along the central axis from coil of N turn is given by:

B(x) = Nμ₀iR²/{2(R² + x²)^3/2}

At point P a distance x = R/2 from each coil, the magnetic field from each coil oint in the +x direction

B_net = B₁ + B₂
B_net = 2[Nμ₀iR²/{2(R² + (R/2)²)^3/2}]
B_net = Nμ₀iR²/(5R²/4)^3/2
       = (200)(4π x 10^─7 Tm/A)(12.2 x 10^─3 A)(0.25 m)²/[5(0.25 m)²/4]^3/2
B_net = (8.78 x 10^─6 T)i

Problem#2
A student makes a short electromagnet by winding 300 turns of wire around a wooden cylinder of diameter d = 5.0 cm. The coil is connected to a battery producing a current of 4.0 A in the wire. (a) What is the magnitude of the magnetic dipole moment of this device? (b) At what axial distance d will the magnetic field have the magnitude 5.0 μT (approximately one-tenth that of Earth’s magnetic field)?

Answer:
(a) the magnitude dipole moment for a coil is given by:

μ = NIA

Where N is the number of turn and A is the cross sectional area of the coil. The area of the coil is A = π(d/2)² where d is the diameter of the coil, hence:

μ = Niπd²/4

subtitute with the givens to get:

μ = 300 x 4.0 A x π x (0.05 m)²/4 = 2.4 A.m²

(b) the magnetic field at distance z along the coils perpendicular central axis is parallel to the axis and is given by:

B = μ₀μ/2πz³

subtitute with the givens to get:

5.0 x 10^─6 T = (4π x 10^─7 T.m/A)(2.4 A.m²)/[2π x z³]
z³ = 0.097336 m³
z = 0.46 m = 46 cm

Problem#3
Figure 3a shows a length of wire carrying a current i and bent into a circular coil of one turn. In Fig.  3b the same length of wire has been bent to give a coil of two turns, each of half the original radius. (a) If B_a and B_b are the magnitudes of the magnetic fields at the centers of the two coils, what is the ratio B_b/B_a? (b) What is the ratio μ_b/μ_a of the dipole moment magnitudes of the coils?

Fig.3

Answer:
The magnitude field a distance x along the central axis from coil is given by:

B(x) = μ₀iR²/{2(R² + x²)^3/2}

(a) At the center of loop x = 0,

B_a = μ₀I/2R

Because 2πR' = ½ (2πR)

R' = ½ R

So that

B_b = μ₀I’/2R’ = μ₀ (2I)/(2 x R/2) = 2μ₀I/R

the ratio B_b/B_a is

B_b/B_a = [2μ₀I/R]/[μ₀I/2R] = 4

(b) the dipole moment magnitudes of the coils

μ = NIA
μ_a = NI(πR²) and
μ_b = N(2I)(πR²/4) = NIπR²/2

therefore:

μ_b/μ_a = 1/2

Problem#4
A 200-turn solenoid having a length of 25 cm and a diameter of 10 cm carries a current of 0.29 A. What is the magnitude of the magnetic dipole moment μ of the solenoid.

Answer:
Known:
The number of turn, N = 200
carries a current, I = 0.29 A
diameter, d = 10 cm

The magnitude of the magnetic dipole moment is given by
μ = NiA
   = 200 x 0.29 A x ¼ π x (0.1 m)²
μ = 0.47 A.m²

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