A solenoid that is 35 cm long and contains 450 circular coils 2.0 cm in diameter carries a 1.75-A current. (a) What is the magnetic field at the centre of the solenoid, 1.0 cm from the coils? (b) Suppose we now stretch out the coils to make a very long wire carrying the same current as before. What is the magnetic field 1.0 cm from the wire’s centre? Is it the same as that in part (a)? Why or why not?

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

Consider a solenoid which is L =35 cm long and contains N= 450 circular coils R =1.0 cm in radius, it carries a current of I =1.75 A. The magnetic field at the center of the solenoid is given by,${\mathrm{B}}_{\text{solenoid}}={\mu }_0\frac{\mathrm{N}}{\mathrm{L}}$ .Substitute the values in the given equation we have,

$\begin{array}{r}{\mathrm{B}}_{\text{solenoid}}=\left(4\pi \times {10}^{-7}\mathrm{Tm}/\mathrm{A}\right)\left(\frac{450}{0.35}\right)\left(1.75\mathrm{A}\right)\\ =2.83\times {10}^{-3}\mathrm{T}\end{array}$

Therefore, the magnetic field at the center of the solenoid is$2.83\times {10}^{-3}\mathrm{T}$

The magnetic field at distance of r=1.0 cm from the center of the wire. Assume that the wire is a very long wire. At distance r from a long straight wire, the magnetic field is$B_{\text{wire}}=\frac{{\mu }_{0}l}{2\pi r}$ Substitute the values in the given equation we have,

$\begin{array}{r}{\mathrm{B}}_{\text{wire}}=\frac{4\pi \times {10}^{-7}\mathrm{Tm}/\mathrm{A}\times 1.75\mathrm{A}}{2\pi \left(1.0\times {10}^{-2}\mathrm{m}\right)}\\ =3.50\times {10}^{-5}\mathrm{T}\end{array}$

Therefore, the magnetic field at distance from the center of the wire is$3.50\times {10}^{-5}\mathrm{T}$