91Ó°ÊÓ

The magnetic field at the center of the solenoid is given by,${\mathit{B}}_{\mathbf{\text{solenoid}}}\mathbf{=}{\mathit{μ}}_{\mathbf{0}}\mathit{n}\mathit{l}$$\mathbf{n}\mathbf{=}\frac{\mathbf{N}}{\mathbf{L}}\mathbf{\text{thus,}}{\mathbf{B}}_{\mathbf{\text{solenoid}}}\mathbf{=}{\mathit{μ}}_{\mathbf{0}}\frac{\mathbf{N}}{\mathbf{L}}\mathbf{\text{I were where,}}{\mathit{μ}}_{\mathbf{0}}\mathbf{=}\mathbf{4}\mathit{π}\mathbf{×}{\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,${\mathrm{B}}_{\text{solenoid}}={\mathrm{μ}}_0\frac{\mathrm{N}}{\mathrm{L}}$ .Substitute the values in the given equation we have,

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

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

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$B_{\text{wire}}=\frac{{\mathrm{μ}}_{0}l}{2\mathrm{π}r}$ Substitute the values in the given equation we have,

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

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