Question

As a new electrical technician, you are designing a large solenoid to produce a uniform 0.150-T magnetic field near the centre of the solenoid. You have enough wire for 4000 circular turns. This solenoid must be 55.0 cm long and 2.80 cm in diameter. What current will you need to produce the necessary field?

Short answer

The current in the magnetic field is 16.4A

Step-by-step solution

Concept of the the magnetic field at the center of the solenoid

The magnetic field at the center of the solenoid is given by $\mathit{B}\mathbf{=}{\mathit{\mu }}_{\mathbf{0}}\mathit{n}\mathit{l}$where ${\mathit{\mu }}_{\mathbf{0}}\mathbf{=}\mathbf{4}\mathit{\tau }\mathit{\tau }\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathit{H}\mathbf{/}\mathit{m}$is permeability of free space, I is the current passing through the coil

STEP 2 Calculate the current in the magnetic field

To produce a uniform magnetic field of B= 0.150 T near the center of the solenoid. We have enough wire for N is 4000 circular turns and this solenoid must be L is 55.0 cm long and R is 1.40 cm in radius. The magnetic field is given as$B={\mu }_{0}nl$ where $n=\frac{N}{L}\text{thus.}B=\frac{{\mu }_{0}NI}{L}$

To find the current in the magnetic field substitute the values in the equation $\mathrm{I}=\frac{\mathrm{BL}}{{\mu }_0\mathrm{N}}$we have,

$\begin{array}{r}\mathrm{I}=\frac{\left(0.150\mathrm{T}\right)\left(0.550\mathrm{m}\right)}{4\pi \times {10}^{-7}\mathrm{Tm}/\mathrm{A}\times 4000}\\ =16.4\mathrm{A}\end{array}$

Therefore, the current in the magnetic field is 16.4A. Also, the magnetic field inside the solenoid is independent of the radius of the solenoid.