Question

A student makes a short electromagnet by winding 540 turns of wire around a wooden cylinder of diameter d = 3.2 cm. The coil is connected to a battery producing a current of 4.3 A in the wire. (a) What is the magnitude of the magnetic dipole moment of this device? (b) At what axial distance z >> d will the magnetic field have the magnitude 5.0 µT (approximately one-tenth that of Earth's magnetic field)?

Answer 1

here given data

turn , N =540

diameter,d= 3.2 cm

d = 0.032 m ( a m = 100 cm )

current , I = 4.3 A

[ a ]

here we want to find magnetic diople moment ,

M = I N A

here I = current

N = turn

A = area

M= magnetic diople moment

here we have to find area,

area = π d² / 4

= 3.1428 * (0.032)² / 4

= 0.0032182272 /4

= 0.0008045568 m²

   A = 80.45 × 10^-5 m²

M = I N A

= 4.3 × 540 ×  80.45 × 10^-5

M = 186804.9 × 10^-5 A·m²

M = 1.868 A·m² , so here magentic moment diople is M = 1.868 A·m²

[b]

here given data

B = 5.0 µT

B = 5 × 10^-6 T ( 1 µT = 10^-6 T )

d = 0.032 m

z = axial distance

here we have general equation for magnetic field of coil with radius r and current I and turn N from the axial distance z,

B = µ₀NIr²/ 2(z² + r²)^3/2 { here radius = r , axial distance = z }

but here z >>> d

so here ( z² + r² ) ^3/2 become ( z² + r² ) ^3/2 = z³

B = µ₀NIr²/ 2(z² + r²)^3/2

B= 4π×10⁻⁷ × 540 × 4.3 × (0.016)² / 2 z³ ( r = d/2 = 0.032/2 = 0.016 m )

5 × 10^-6 = 7.4727 ×10⁻⁷ / 2 z³

2 z³ = 7.4727×10⁻⁷ / 5 × 10^-6

2 z³ = 1.4945 ×10^-1

z³ = 0.14945 / 2

z³ = 0.7472

z = (0.7472 )^1/3

z = 0.9074 m ,

so here at axial distance z = 0.9074 m magnetic field is  5.0 µT