contestada






Exercise 28.41

Two round concentric metal wires lie on a tabletop, one inside the other. The inner wire has a diameter of 24.0cm and carries a clockwise current of 10.0A , as viewed from above, and the outer wire has a diameter of 38.0cm .



Part A

What must be the direction (as viewed from above) of the current in the outer wire so that the net magnetic field due to this combination of wires is zero at the common center of the wires?

The currents direction must be clockwise

or

The currents direction must be counterclockwise

Part B

What must be the magnitude of the current in




Exercise 28.41

Two round concentric metal wires lie on a tabletop, one inside the other. The inner wire has a diameter of 24.0cm and carries a clockwise current of 10.0A , as viewed from above, and the outer wire has a diameter of 38.0cm .



Part A

What must be the direction (as viewed from above) of the current in the outer wire so that the net magnetic field due to this combination of wires is zero at the common center of the wires?

The currents direction must be clockwise

or

The currents direction must be counterclockwise

Part B

What must be the magnitude of the current in the outer wire so that the net magnetic field due to this combination of wires is zero at the common center of the wires?


















I=


A

Respuesta :

MissPhiladelphia
Part A.
In order for the net magnetic field at the center of the circles to be zero, the individual magnetic fields produced by the wires must cancel each other. So in terms of direction, they must be opposite. The answer is:
The currents direction must be counterclockwise

Part B
We have the relationship:
I₁/D₁ = I₂/D₂

10/24 = I₂/38

I₂ = 15.83 A
onyebuchinnaji

(a) The currents direction must be counterclockwise.

(b) The magnitude of the current in the outer wire is 15.83 A.

Direction of the current

The direction of the current will flow in such a way that the magnetic field due to the wires combination will cancel out. Thus, the current will flow in opposite or counterclockwise direction.

Magnitude of the current

The magnitude of the current is calculated using the following formulas;

[tex]\frac{I_1}{D_1} = \frac{I_2}{D_2} \\\\I_2 = \frac{I_1 D_2}{D_1} \\\\I_2 = \frac{10 \times 38}{24} \\\\I_2 = 15.83 \ A[/tex]

Thus, the magnitude of the current in the outer wire is 15.83 A.

Learn more about current in outer wire for zero electric field here: https://brainly.com/question/13200775

Otras preguntas