Question

A student makes a short electromagnet by winding 300 turns of wire around a wooden cylinder of diameter d=5cm. 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 $z>>d$ will the magnetic field have the magnitude $5.0\mu T$ (approximately one-tenth that ofEarth’s magnetic field)?

Answer 1

a)The magnitude of the magnetic dipole moment is given by $\mu =Nia$, where $N$ is the number of turns, $i$ is the current and $A$ is the area. We use $A=\pi R^2$, where $R$ is the radius. Thus,
$\mu =Ni\pi R^2=\left(300\right)\left(4.0A\right)\pi \left(0.025m\right)=2.4A.m^2$.
(b)The magnetic field on the axis of a magnetic dipole, a distance z away, $B=\frac{{\mu }_0}{2\pi }\frac{\mu }{z^3}$.
We solve for z:
$z=(\frac{{\mu }_0}{2\pi }\frac{\mu }{B}{)}^{1/3}=(\frac{(4\pi \times {10}^{-7}T.m/A)(2.36A.m^2)}{2\pi (5.0\times {10}^{-6}T)}{)}^{1/3}=46cm$