Answer:
I=1.48 A
Explanation:
Given that
B=3.1 x 10⁻5 T
b= 4.2 cm
l= 9.5 cm
The relationship for magnetic field and current given as
[tex]B=\dfrac{2\mu _oI}{\pi}D[/tex]
Where
[tex]D=\dfrac{\sqrt{l^2+b^2}}{lb}[/tex]
By putting the values
[tex]D=\dfrac{\sqrt{l^2+b^2}}{lb}[/tex]
[tex]D=\dfrac{\sqrt{0.042^2+0.095^2}}{0.042\times 0.095}[/tex]
D=26.03 m⁻¹
[tex]B=\dfrac{2\mu _oI}{\pi}D[/tex]
[tex]3.1\times 10^{-5}=\dfrac{2\times 4\times \pi \times 10^{-7} I}{\pi}\times 26.03[/tex]
[tex]I=\dfrac{3.1\times 10^{-5}}{{2\times 4\times 10^{-7} }\times 26.03}[/tex]
I=1.48 A
