Answer:
<em>The direction of the magnetic field on point P, equidistant from both wires, and having equal magnitude of current flowing through them will be pointed perpendicularly away from the direction of the wires.</em>
Explanation:
Using the right hand grip, the direction of the magnet field on the wire M is counterclockwise, and the direction of the magnetic field on wire N is clockwise. Using this ideas, we can see that the magnetic flux of both field due to the currents of the same magnitude through both wires, acting on a particle P equidistant from both wires will act in a direction perpendicularly away from both wires.
Answer:
When there is a change in magnetic flux linkage through a loop of wire, an electromotive force is induced in the loop, according to the Faraday-Newmann-Lenz Law:

where
N is the number of turns in the loop
is the change in magnetic flux through the loop
is the time elapsed
The negative sign in the formula represents Lenz's Law, and tells us about the direction of the electromotive force.
In fact, the negative sign means that the direction of the induced emf is such that to oppose to the change in the magnetic flux that originated the induced emf.
This is a consequence of the law of conservation of energy: no energy can be created out of nowhere. In fact, when the emf is induced in the loop, electrical energy appears in the circuit; however, this electric energy cannot come out of nowhere. Instead, it is just "created" from the transformation of some other form of energy (for instance, the mechanical energy that is used to move the loop in the magnetic field, and changing its magnetic flux).
The negative sign in Lenz's Law tells exactly this: the direction of the induced emf is such that it opposes the initial change in magnetic flux that generated the induced emf, so that overall the total energy is conserved.
The answer is orbit, we are orbiting the sun as the moon orbits us
Answer:
M g H / 2 = M g L / 2 initial potential energy of rod
I ω^2 / 2 = 1/3 M L^2 * ω^2 / 2 kinetic energy attained by rod
M g L / 2 = 1/3 M L^2 * ω^2 / 2
g = 3 L ω^2
ω = (g / (3 L))^1/2
Work done on the crate is 1411.2 J
Explanation:
Work done is defined as the product of force and the distance moved by the object. The unit of work done is in joules and denoted by the symbol J.
Work done = F * d
where F represents the force and d represents the distance moved by the object.
mass = 72 kg , distance moved by the object is given by 2.0 m
Force F = mass * gravity = 72 * 9.8
= 705.6 N =706 N.
Work done = 706 * 2.0 = 1412 J.