A segment of wire carries a current of 25 A along the x axis from x = −2.0 m to x = 0 and then along the y axis from y = 0 to y
= 3.0 m. In this region of space, the magnetic field is equal to 40 mT in the positive z direction. What is the magnitude of the force on this segment of wire?
1 answer:
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The tralational equilibrium condition allows finding that the electric potential is V = 4.8 10¹¹ V
Given parameter
- The mass m = 1.5 g = 1.15 10-3 kg
- The charge on the sphere q = 8.9 10-16 C
- Plate spacing d = 5 cm = 5.00 10-2 m
To find
- The potential difference
Newton's second law states that the force is proportional to the mass and the acceleration of the bodies, in the special case that the acceleration is zero, it establishes the condition for the equilibrium of the bodies
∑ F = 0
Where the bold indicate vector and F is the force
To use this equation we must fix a reference system with respect to which to carry out the decomposition and measurements of the forces; let's fix a system with the horizontal x axis and the vertical y axis, in the attachment I could see a free body diagram.
x- axis
Fe - Tₓ = 0
Fe = Tₓ
y-axis
- W = 0
W =
mg =
The electric force is
Fe = q E = q V / d
let's use trigonometry to decompose the stress
cos 30 = / T
sin 30 = Tₓ / T
= T cos 30
Tₓ = T sin 30
We substitute
q V / d = T sin 30
mg = T cos 30
It's solve the system of equations
= tan 30
V =
It's calculate
V =
V = 4.768 10¹¹ V
In conclusion, using the equilibrium condition, we could find that the electric potential is V = 4.8 10¹¹ V
Learn more about equilibrium condition here:
