a uniform magnetic field of strength 0.5T is directed at an angle of 30degree to the plane of 100turns regular copper coil of lenght 0.04m and width 0.05mm. the diameter of the copper wire which makes up a coil is 1mm and the resistivity of copper is 0.000000017ohms meter. the magnetic field through the loop decreases to 0 in 0.3seconds. determine
1) initial coil through the coil
2) the average induced E.M.F in the coil as the field increases
3) the amount of charge that passes the coil as the field increases
To answer these questions, we'll use Faraday's law of electromagnetic induction and also consider the given properties of the copper coil.
1) To find the initial magnetic flux through the coil, we can use the equation:
Φ = B * A * cos(θ)
Where:
Φ is the magnetic flux
B is the magnetic field strength
A is the area of the coil
θ is the angle between the magnetic field and the plane of the coil
Given:
B = 0.5 T
A = length * width = 0.04m * 0.05mm = 0.000002m^2 (Note: we converted mm to m)
θ = 30°
Substituting the given values into the equation:
Φ = 0.5 T * 0.000002 m^2 * cos(30°)
Simplifying:
Φ = 0.5 * 0.000002 * 0.866
Φ = 0.000000000866 Wb
Therefore, the initial magnetic flux through the coil is 0.000000000866 Weber.
2) The average induced electromotive force (EMF) in the coil can be calculated using the equation:
E = -(ΔΦ/Δt)
Where:
E is the EMF induced
ΔΦ is the change in magnetic flux
Δt is the change in time
Given:
ΔΦ = 0.000000000866 Wb (final magnetic flux is 0, so change is the initial flux)
Δt = 0.3 seconds
Substituting the values:
E = - (0.000000000866 Wb / 0.3 s)
E ≈ -0.00000000289 V
Therefore, the average induced EMF in the coil as the field increases is approximately -0.00000000289 Volts.
Note: The negative sign indicates that the direction of the induced EMF opposes the change in magnetic flux.
3) To find the amount of charge that passes the coil as the field increases, we need to use the equation:
Q = I * t
Where:
Q is the charge
I is the electric current
t is the time
Given:
t = 0.3 seconds (as mentioned in question)
R = resistivity of copper = 0.000000017 Ω·m
l = length of the wire = 100 turns * 0.04 m
A = cross-sectional area of the wire = π * (diameter / 2)^2
Given the diameter of the wire is 1mm, and we need to convert it to meters:
d = 1mm * 0.001m/mm = 0.001m
Now, substituting these values into the equations:
l = 100 * 0.04 m = 4 m
A = π * (0.001m/2)^2 = 7.854×10^-7 m^2
To find the current, we can use Ohm's Law:
V = IR
R = ρ * (l / A)
Substituting the given values:
R = 0.000000017 Ω·m * (4 m / 7.854×10^-7 m^2)
R ≈ 0.0862 Ω
Now, find the current:
V = IR
I = V / R
Since V = E (from part 2), we can use the value of E to find I:
I = -0.00000000289 V / 0.0862 Ω
I ≈ -3.36×10^-8 A
Now, substitute the values of I and t into the equation for charge:
Q = I * t
Q = -3.36×10^-8 A * 0.3 s
Q ≈ -1.008×10^-8 C
Therefore, the amount of charge that passes the coil as the field increases is approximately -1.008×10^-8 Coulombs.