Question

A student makes a short electromagnet by winding 480 turns of wire around a wooden cylinder of diameter d = 3.5 cm. The coil is connected to a battery producing a current of 4.8 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 4.7 µT (approximately one-tenth that of Earth's magnetic field)?

(a)

Number

Enter your answer for part (a) in accordance to the question statement

Units

Choose the answer for part (a) from the menu in accordance to the question statement CC·mC/m^3C/m^2C/mnCmCμCAA·m^2A/m^2mAA/mA/sN·m^2/CpC

(b)

Number

Enter your answer for part (b) in accordance to the question statement

Units

Choose the answer for part (b) from the menu in accordance to the question statement This answer has no units° (degrees)mkgsm/sm/s^2NJWN/mkg·m/s or N·sN/m^2 or Pakg/m^3gm/s^3times

Answer 1

Diameter of a wooden cylinder, d = 3.5 cm = 3.5 x 10^-2 m

Radius of a wooden cylinder, r = 1.75 x 10^-2 m

(a) What is the magnitude of a magnetic dipole moment of this device?

_B = N I A

where, N = number of turns = 480

I = current flowing in the wire = 4.8 A

A = cross-sectional area of the coil = r²

then, we get

_B = [(480) (4.8 A) (3.14) (1.75 x 10^-2 m)²]

_B = 2.21 A.m²

(b) At what axial distance z >> d, will the magnetic field have a magnitude 4.7 T?

We know that, the magnetic field at distance (z) along the coil's perpendicular central axis is parallel to the axis.

Then, we have

B = (₀ / 2) (_B / z³)

Rearranging an above eq. & we get

z = (_0B / 2B)^1/3

z = {[(4 x 10^-7 mT/A) (2.21 A.m²)] / [(2) (4.7 x 10^-6 T)]}^1/3

z = 0.454 m

Converting m into cm :

z = 45.4 m