It is a false statement i.e. drift velocity is not same in the direction as the applied force.
Drift velocity of a current-carrying conductor can be explained as, the charges i.e. electrons do not flow in the same direction of current. In other word, in most cases the movement of the electrons is almost random, with a small net velocity. So that , the drift velocity, in the direction opposite to the electric field.
Drift velocity 
is inversely proportional to the number of electron per unit volume of the conductor e. Therefore, the formulation can be given as ,
= σ E/ne
The above equation shows the drift velocity in a current carrying conductor
where, is drift velocity , σ is the conductivity, E is electric force and n is number of electrons per unit volume of the conductor e.
Hence here we can say that, the drift velocity is not in the same direction as the applied force.
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All o the galaxies are moving in a same speed
Answer:
The electric current in the wire is 0.8 A
Explanation:
We solve this problem by applying the formula of the magnetic field generated at a distance by a long and straight conductor wire that carries electric current, as follows:
B= Magnetic field due to a straight and long wire that carries current
u= Free space permeability
I= Electrical current passing through the wire
a = Perpendicular distance from the wire to the point where the magnetic field is located
Magnetic Field Calculation
We cleared (I) of the formula (1):
Formula(2)
a =8cm=0.08m
We replace the known information in the formula (2)
I=0.8 A
Answer: The electric current in the wire is 0.8 A
Answer: 288.8 m
Explanation:
We have the following data:
is the time it takes to the child to reach the bottom of the slope
is the initial velocity (the child started from rest)
is the angle of the slope
is the length of the slope
Now, the Force exerted on the sled along the ramp is:
(1)
Where 
is the mass of the sled and 
its acceleration
In addition, if we draw a free body diagram of this sled, the force along the ramp will be:
(2)
Where 
is the acceleration due gravity
Then:
(3)
Finding :
(4)
(5)
(6)
Now, we will use the following kinematic equations to find :
(7)
(8)
Where 
is the final velocity
Finding from (7):
(9)
(10)
Substituting (10) in (8):
(11)
Finding :
