Respuesta :
Answer: The outer loop should carry a current of 2.0 A.
The current should flow in the counter-clockwise direction
Explanation: Please see the attachments below
![Ver imagen Abdulazeez10](https://us-static.z-dn.net/files/dc6/4d5cc921d3c506c314768a755ec8b30a.jpg)
![Ver imagen Abdulazeez10](https://us-static.z-dn.net/files/d33/43c3072f7ea8b366e2bdaf9f92a757d8.jpg)
Using the relation for the the current in a magnetic loop, the current and direction in the inner loop would be 2.0 amperes in the clockwise direction
Using the relation for the magnetic field in a loop :
- [tex] B =\frac{U_{0}I}{2R}[/tex]
- R = Radius ; I = current
- Inner loop Radius = [tex] R_{0} [/tex]
- Outer loop Radius = [tex] R_{0} = 2R_{1} [/tex]
outer loop = Inner loop
[tex] \frac{I_{0}}{2 \times 2 R_{1}} = \frac{I_{1}}{2R_{1}}[/tex]
Current of inner loop = 1.0 ampere
[tex] \frac{I_{0}}{4R_{1}} = \frac{1}{2R_{1}}[/tex]
Cross multiply :
[tex] I_{0} \times 2R_{1} = 4R_{1}[/tex]
[tex] I_{0} = \frac{4R_{1}}{2R_{1}}[/tex]
[tex] I_{0} = 2[/tex]
Hence, the current of the inner loop is 2.0 ampere and it is in the clockwise direction.
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