# 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

#### Answer:

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

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