Answer:
The magnitude of the magnetic torque on the loop when the plane of its area is perpendicular to the magnetic field is 0.4713 J
Explanation:
Given;
radius of the circular loop of wire = 0.5 m
current in circular loop of wire = 2 A
strength of magnetic field in the wire = 0.3 T
τ = μ x Bsinθ
where;
τ is the magnitude of the magnetic torque
μ is the dipole moment of the magnetic field
θ is the inclination angle, for a plane area perpendicular to the magnetic field, θ = 90
μ = IA
where;
I is current in circular loop of wire
A is area of the circular loop = πr² = π(0.5)² = 0.7855 m²
μ = 2 x 0.7885 = 1.571 A.m²
τ = μ x Bsinθ = 1.571 x 0.3 sin(90)
τ = 0.4713 J
Therefore, the magnitude of the magnetic torque on the loop when the plane of its area is perpendicular to the magnetic field is 0.4713 J
Each dot represents a valence electron. Valence electrons are the electron on the outer electron shell of an atom.<span>
</span>
What’s the question? Is it true or false?
Missing question: "What is the spring's constant?"
Solution:
The object of mass m=6.89 kg exerts a force on the spring equal to its weight:

When the object is attached to the spring, the displacement of the spring with respect to its equilibrium position is

And by using Hook's law, we can find the constant of the spring: