The magnetic field strength at the north pole of a $2.0cm$- diameter, $8cm$-long Alnico magnet is $0.10T$. To produce the same field with a solenoid of the same size, carrying a current of $2.0A$, how many turns of wire would you need?

We need to find the turns of wire would need.

The solenoid is wounded in a number of turns$N$and due to the current in this wounded wire, here is a magnetic field produced in a uniform shape at the center of the solenoid where equation (29.17) gives the magnetic field everywhere inside the infinite solenoid and along the central axis of an infinite solenoid the magnetic field is given as:

localid="1650882874317" $B=\frac{{\mu }_{0}NI}{I}$

Where $N$is the number of turns, $L$its length, $I$is the current in the solenoid and ${\mu }_0$is the free space permeability and equals

$4\pi \times {10}^{-7}T·m/A·$

By solving for $N$as:

$N=\frac{Bl}{{\mu }_{o}I}$

We want to produce a magnetic field of $B=0.10T$with the same length of $L=8cm$.So, let us plug the values for ${\mu }_0$$B,LandI$into equation (2) to get $N$:

$$\begin{gathered} N=\frac{Bl}{{\mu }_{o}I} \\ =\frac{(0.10T)(0.08m)}{(4\pi \times {10}^{-7}T·m/A)(2A)} \\ =\text{3185turns} \end{gathered}$$