Answer:
The width of the sheet is [tex]w =0.8046m[/tex]
Explanation:
From the question we are told that
The distance of point P above a large sheet of metal is [tex]D = 4.0mm =\frac{4}{1000} = 0.004m[/tex]
The current on the large metal sheet is [tex]I =34A[/tex]
The distance of the the point P below a long wire [tex]d = 3.0mm = \frac{3}{1000} = 0.003m[/tex]
The current on the long wire is [tex]I_w = 0.41A[/tex]
The magnetic field at P is [tex]B = 0T[/tex]
Generally magnetic field of P long wire is mathematically represented as
[tex]B_w = \frac{\mu_o I_w}{2\pi r}[/tex]
Generally magnetic field of P large sheet of meta is mathematically represented as
[tex]B_m = \frac{\mu_o K}{2}[/tex]
Where K is the current per unit width
The total magnetic field at P is
[tex]\frac{\mu_o I_w}{2 \pi r} = \frac{\mu_o K}{2}[/tex]
Making K the subject of formula
[tex]K = \frac{2 I_w }{2 \pi r }[/tex]
Substituting values
[tex]K = \frac{2 * 0.41 }{2 * 3.142 * (0.0030) }[/tex]
[tex]K = 43.4967 A/m[/tex]
Generally K is mathematically represented as
[tex]K = \frac{I}{w}[/tex]
Where w is the width of the large sheet
Therefore the width of the metal sheet [tex]w = \frac{I}{K}[/tex]
[tex]= \frac{35}{43.4967}[/tex]
[tex]w =0.8046m[/tex]
