# Magnetic field strength in a solenoid

I'm pretty new to physics. I've been conducting some experiments with electromagnets. My practical results don't match up with the theory.

The magnetic field in a solenoid of length $L$ around an iron core with $N$ turns is given by: $$B = \mu \frac{NI}{L}.$$

Assuming Ohm's law of resistance in the wire we can replace $I$ with $V/R$ to get $$B = \mu \frac{NV}{LR}.$$

In my experiments (using batteries and copper wire around a ferrite core), doubling the number of turns in the solenoid does gives approximately (although slightly less than, as expected) double the lifting strength of my magnet.

However doubling the voltage (1.5V to 3V by connecting batteries in serial) seems to only give an increased lifting strength of about 50%.

Any ideas as to why?

Many thanks

-

If $B \propto I$, then by Ohm's Law $B \propto {V \over R}$
If the internal resistance of dry cells are large compared to solenoid wire resistance, then doubling the number of batteries will not double the magnetic flux through the solenoid since the resistance of the circuit increases accordingly. (It is not true that $no. of ~batteries \propto I$)
But if the wire resistance is small compared to battery internal resistance, then doubling the number of coils negligibly affects the resistance of the circuit, in which case you may in fact use $B\propto NI$ directly (where $I$ is constant).
Finally as you would have noticed I have not referred to magnetic force but magnetic field. Note that the magnetic force $F\propto B^2$, thereby in your case you should be expecting $F\propto N^2$, a quadrupling of holding force when number of loops are doubled. The reason why you see a linear relationship instead is due to the relative permeability of the iron core $\mu_r$ ($\mu=\mu_0\mu_r$) being variable to the magnetic fields originating in the nearby solenoid. therefore you wont see a $F\propto B^2$ until the iron core is saturated, i.e, use higher currents. Even if you used a precise voltmeter across the solenoid so you can use the first equation correctly (to avoid the assumption that current is proportional to number of batteries), the variable $\mu_r$ will still affect the results.