Question

A solenoid is wound with 259 turns per cm. An outer layer of insulated wire with 51 turns per cm is wound over the first layer of wire. The inner coil carries a current of 7.577 A, and the outer coil carries a current of 21.68 A in the opposite direction. What is the magnitude of the magnetic field, in microTeslas, at the central axis ?

Answer 1

Answer tab

For this question, we'll first define a number, called the "linear density of the coil" which is exactly the one the exercise gives us, in a unit of turns per length. The magnetic field can be calculated as:

`B=\mu_0in`

Where n is the linear density.

In our case, as the coils carry current in opposite directions, the generated magnetic fields will be opposed, and we'll have:

`B=B_(outer)-B_(inner)=\mu_0*(51*10^2*21.68-259*10^2*7.577)`

Please note that we had to multiply by 10^2, in order to convert turns/cm to turns/m

Then, our final magnetic field will be:

`|B|=0.107664T=107664\mu T`

Our final answer is B=107664uT