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
The acceleration of an object depends directly upon the net force acting upon the object, and inversely upon the mass of the object. As the force acting upon an object is increased, the acceleration of the object is increased. As the mass of an object is increased, the acceleration of the object is decreased.
Change in velocity of larger moose: (1/3)v - v = -(2/3)v
<span>change in velocity of small moose: (1/3)v - (-v) = (4/3)v </span>
<span>- (change in velocity of larger moose)/(change in velocity of smaller moose) = 2
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
</span>
sign out and log in again...if does not work then make a new account
Answer:
q=3.5*10^-4
Explanation:
<u>concept:</u>
The force acting on both charges is given by the coulomb law:
F=kq1q2/r^2
the centripetal force is given by:
Fc=mv^2/r
The kinetic energy is given by:
KE=1/2mv^2
<u>The tension force:</u>
<u><em>when the plane is uncharged </em></u>
T=mv^2/r
T=2(K.E)/r
T=2(50 J)/r
T=100/r
<u><em>when the plane is charged </em></u>
T+k*|q|^2/r^2=2(K.E)charged/r
100/r+k*|q|^2/r^2=2(53.5 J)/r
q=√(2r[53.5 J-50 J]/k) √= square root on whole
q=√2(2)(53.5 J-50 J)/8.99*10^9
q=3.5*10^-4
