Question

A student makes a short electromagnet by winding 340 turns of wire around a wooden cylinder of diameter d = 2.1 cm. The coil is connected to a battery producing a current of 4.4 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 7.4 µT (approximately one-tenth that of Earth's magnetic field)?

Answer 1

Step-by-step explanation:

Magnetic moment of one turn of current carrying coil = current x area of coil

= 4.4 x π x 1.05² x 10⁻⁴ Am²

magnetic moment of 340 turns

M = 340 x 4.4 x π x 1.05² x 10⁻⁴

= .5179 Am²

This will act as short bar magnet . Magnetic field due to a short bar magnet at a point on its axis is given by following formula

B = 2 k M / d³ , k is a constant

Putting the given value of magnetic field we can find d.

7.4 x 10⁻⁶ = 2 x 4π x 10⁻⁷ x .5179 / d³

d³ = 2 x 4π x 10⁻⁷ x .5179 / 7.4 x 10⁻⁶

= .1758 m

d = .56 m

= 56 cm