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Sagot :
Answer:
1.63 A and in clockwise direction
Explanation:
The magnetic field due to the rectangular loop is :
[tex]$B=\frac{2 \mu_0 I}{\pi}\left(\frac{\sqrt{L^2+W^2}}{LW}\right)$[/tex]
Given : W = 4.20 cm
[tex]$=4.20 \times 10^{-2} \ m$[/tex]
L = 9.50 cm
[tex]$= 9.50 \times 10^{-2} \ m$[/tex]
[tex]$B = 3.40 \times 10^{-5} \ T $[/tex]
Rearranging the above equation, we get
[tex]$I=\frac{B \pi LW}{2 \mu_0\sqrt{L^2+W^2}}$[/tex]
[tex]$I=\frac{(3.40 \times 10^{-5}) \pi(9.50 \times 10^{-2})(4.20 \times 10^{-2})}{2(4 \pi \times 10^{-7})\sqrt{(9.50 \times 10^{-2})^2+(4.20 \times 10^{-2})^2}}$[/tex]
I = 1.63 A
So the magnitude of the current in the rectangular loop is 1.63 A.
And the direction of current is clockwise.
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