If the wire were wrapped around nickel instead of the iron core and the original current were applied, how strong would the magnetic field be?

An electromagnet is created by wrapping 28 turns of copper wire around a 65cm long iron core. A current of 800mA is passed through the wire.

Mar 1, 2016

$= 4.33 \times {10}^{-} 3 T$

Explanation:

The magnetic field strength $H$, inside free space of a solenoid having number of turns per unit length ${N}_{0}$, carrying a current $I$ is calculated as follows:

We know that $H$ is directly proportional to number of turns per unit length.
Also that it is directly proportional to the current flowing through it. mathematically
$H \propto {N}_{0}$
$H \propto I$
Combining two we obtain
$H \propto {N}_{0} \cdot I$

or $H = {\mu}_{\circ} {N}_{0} \cdot I$
where ${\mu}_{\circ}$ is a constant of proportionality and is called magnetic permeability of free space and has a value $4 \pi \times {10}^{-} 7 N {A}^{-} 2$
For the problem
$H = 4 \pi \times {10}^{-} 7 \cdot \frac{28}{0.65} \cdot 0.8$
$H = 4.33 \times {10}^{-} 5 T$

It is given that core of the solenoid is of Nickel. Hence, magnetic field produced in the core is

$B = {\mu}_{r} \times H$, where ${\mu}_{r}$ is the relative permeability of nickel.

Recall the permeability curves. Permeability of nickel varies from 100 to 600 maximum.
Since the order of magnitude of magnetizing field $H$ is around ${10}^{-} 5$, it would be safe to use lower value of relative permeability for the nickel core.

$\therefore B = 100 \times 4.33 \times {10}^{-} 5$
or $= 4.33 \times {10}^{-} 3 T$