Answer:
The Flemings left hand rule is used to find the magnitude of a magnetic force
Explanation:
Fleming's left hand rule states that if the first three fingers are held mutually at right angles to one another, then the fore finger points into the direction of magnetic field the middle finger in the direction of current while the thumb points in the direction of force.
Mathematically
Magnetic Force F= BILsinθ
Where
B= magnetic field density Tesla
I= current
L= length of conductor
θ= angle of conductor with field
Answer: 8*10^-15 N
Explanation: In order to calculate the force applied on an electron in the middle of the two planes at 500 V we know that, F=q*E
The electric field between the plates is given by:
E = ΔV/d = 500 V/0.01 m=5*10^3 N/C
the force applied to the electron is: F=e*E=8*10^-15 N
Answer:

Explanation:
It is given that,
Length of the wire, L = 0.6 m
Current flowing inside the wire, I = 2 A
Uniform magnetic field, B = 0.3 T
Force experienced by the wire in the magnetic field, F = 0.18 N
To find,
The angle made by the wire with the magnetic field.
Solve,
We know that the magnetic force acting on the wire inside the magnetic field is given by :




Therefore, the wire makes an angle of 30 degrees with respect to magnetic field.
Answer:
a) a = 4.9 m / s², N = 16.97 N and b) F = 9.8 N
Explanation:
a) For this exercise we will use Newton's second law, we write a reference system with the x axis parallel to the plane, see attached, in this system the only force we have to break down is weight, let's use trigonometry
sin 30 = Wx / W
cos 30 = Wy / W
Wx = W sin30
Wy = W cos 30
Let's write the equations on each axis
X axis
Wx = ma
Y Axis
N- Wy = 0
N = Wy = mg cos 30
N = 2.0 9.8 cos 30
N = 16.97 N
We calculate the acceleration
a = Wx / m
a = mg sin 30 / m
a = g sin 30
a =9.8 sin 30
a = 4.9 m / s²
b) For the block to move with constant speed, the acceleration must be zero, so the force applied must be equal to the weight component
F -Wx = 0
F = Wx
F = m g sin 30
F = 2.0 9.8 sin 30
F = 9.8 N
Answer: im not gonna give i to you just do 15+15=_+ 5.6+6.4 easy
Explanation: i took the test and got a 100%