Answer:
The magnitude of the magnetic field B at the center of the loop is 5.0272 x 10⁻⁴ T.
Explanation:
Given;
Radius of circular loop, R = 3.00 cm = 0.03 m
Current in the loop, I = 12.0 A
Magnetic field at the center of circular loop is given as;
B = μ₀I / 2R
Where;
μ₀ is constant = 4π x 10⁻⁷ T.m/A
R is the radius of the circular loop
I is the current in the loop
Substitute the given values in the above equation and calculate the magnitude of the magnetic field;
B = (4π x 10⁻⁷ x 12)/ 0.03
B = 5.0272 x 10⁻⁴ T
Therefore, the magnitude of the magnetic field B at the center of the loop is 5.0272 x 10⁻⁴ T.
We can solve the problem by using the first law of thermodynamics:
where
is the variation of internal energy of the system
Q is the heat added to the system
W is the work done by the system
In this problem, the variation of internal energy of the system is
While the heat added to the system is
therefore, the work done by the system is
"(1) a satellite moving around Earth in a circular <span>orbit" is the only option from the list that describes an object in equilibrium, since velocity and gravity are working together to keep the orbit constant. </span>
B. The velocity of the second flight is negative compared to the speed.
