A straight 1.20 m long conductor has a 2.00 A current travelling toward the East. Earth's magnetic field in this location is 4.9
1 answer:
Answer:
Force's magnitude
Direction: down (towards the center of the Earth)
Explanation:
Recall that the magnetic force on a conductor of length L carrying a current I in a magnetic field B is given by the equation: in the case the magnetic field B and the direction of the current are at 90 degrees from each other (which is our case). The direction of the force will be given by the "right hand rule" associated with the vector product that defines this force.
Since the current is moving East, and the magnetic field of the Earth goes from North to South, the resultant Force vector will be pointing towards the Earth (and perpendicular to the plane defined by the current's direction and the magnetic field B)
The magnitude of the force, is given by the formula above, and given that all quantities to be considered are is SI units, it will result in Newtons (N):
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Answer:
Explanation:
1 )
We shall apply conservation of momentum law to solve the problem.
mv = ( M +m) V , m and M are masses of small and large object , v is the velocity of small object before collision and V is the velocity of both the objects together after collision .
.5 x .2 = (1.5 + .5)V
V = .05 m /s
2 ) We shall use formula for velocity of object after elastic collision as follows
v₁ =
m₁ and m₂ are masses of first and second object u₁ and u₂ are their initial velocity and v₁ and v₂ are their final velocity.
Putting the values
=
= - .66 m /s
Since the sign is negative so it will be in opposite direction .