A 5.0 cm diameter coil of 20 turns with a resistance of 0.50 Ù is in a time-varying magnetic field which is perpendicular to the coil and is given by the expression;
B = 0.010t2 + 0.020t, where B is in tesla and t is in seconds.
a) Draw a graph of B as a function of time from t = 0 s to t = 10 s.
b) Write down an expression for the induced current I(t) as a function of time.
c) Evaluate I at t = 5 s and t = 10 s.
a) I'd love to help you draw a graph, but I'm just a Clown Bot, and my art skills are limited to balloon animals. Maybe you could find a helpful human to assist you with that!
b) To write down the expression for the induced current, we can use Faraday's law of electromagnetic induction. The induced electromotive force (emf) in a coil is given by the equation:
emf = -N * dΦ/dt
Where N is the number of turns in the coil and dΦ/dt is the rate of change of magnetic flux through the coil. In this case, the magnetic flux is given by:
Φ = B * A
Where B is the magnetic field and A is the area of the coil. The area can be calculated from the diameter:
A = π * (d/2)^2
Putting it all together, the induced current can be expressed as:
I(t) = -N * (dΦ/dt) / R
Where R is the resistance of the coil.
c) To evaluate I at t = 5s and t = 10s, you'll need to substitute these values into the expression for I(t) that we derived in part (b).
a) To draw the graph of B as a function of time, we can plot points by substituting different values of t into the expression B = 0.010t^2 + 0.020t. We'll use values of t from 0 to 10 seconds.
t = 0 s:
B = 0.010(0)^2 + 0.020(0) = 0 T
t = 1 s:
B = 0.010(1)^2 + 0.020(1) = 0.030 T
t = 2 s:
B = 0.010(2)^2 + 0.020(2) = 0.080 T
t = 3 s:
B = 0.010(3)^2 + 0.020(3) = 0.150 T
t = 4 s:
B = 0.010(4)^2 + 0.020(4) = 0.240 T
t = 5 s:
B = 0.010(5)^2 + 0.020(5) = 0.350 T
t = 6 s:
B = 0.010(6)^2 + 0.020(6) = 0.480 T
t = 7 s:
B = 0.010(7)^2 + 0.020(7) = 0.630 T
t = 8 s:
B = 0.010(8)^2 + 0.020(8) = 0.800 T
t = 9 s:
B = 0.010(9)^2 + 0.020(9) = 0.990 T
t = 10 s:
B = 0.010(10)^2 + 0.020(10) = 1.200 T
We can now plot these points on a graph with the x-axis representing time (t) from 0 to 10 seconds, and the y-axis representing magnetic field (B) from 0 to 1.5 Tesla.
b) To find the expression for the induced current I(t) as a function of time, we can use Faraday's law of electromagnetic induction:
ε = -dΦ/dt
Where ε is the induced electromotive force (emf), Φ is the magnetic flux through the coil, and t is time.
Given that the emf ε is equal to the product of the induced current I(t) and the resistance R, we can write:
ε = -dΦ/dt = -N d(BA)/dt
Where N is the number of turns in the coil, B is the magnetic field, and A is the area of the coil.
For a circular coil, A = πr^2, where r is the radius of the coil.
Differentiating B = 0.010t^2 + 0.020t with respect to t, we get:
dB/dt = 0.020t + 0.020
Substituting this into the expression for ε:
ε = -N d(BA)/dt = -N A d(B)/dt = -N(A)(0.020t + 0.020)
The emf ε is equal to I(t) * R, so we have:
I(t) * R = -N(A)(0.020t + 0.020)
Solving for I(t):
I(t) = (-N(A)(0.020t + 0.020)) / R
Given that the radius r = 5 cm = 0.05 m, the number of turns N = 20, and the resistance R = 0.50 Ω, we can substitute these values into the expression to get a final expression for I(t).
c) To evaluate I at t = 5 s and t = 10 s, we simply substitute these values of t into the equation for I(t):
At t = 5 s:
I(5) = (-20(π(0.05)^2)(0.020(5) + 0.020)) / 0.50
At t = 10 s:
I(10) = (-20(π(0.05)^2)(0.020(10) + 0.020)) / 0.50
a) To draw a graph of B as a function of time, we can substitute different values of t into the given expression for B and plot the corresponding values on a graph.
Let's consider values of t in the range from 0s to 10s. We'll calculate the corresponding values of B using the expression B = 0.010t^2 + 0.020t.
For example, when t = 0s, B = 0.010(0)^2 + 0.020(0) = 0 Tesla.
When t = 1s, B = 0.010(1)^2 + 0.020(1) = 0.030 Tesla.
When t = 2s, B = 0.010(2)^2 + 0.020(2) = 0.080 Tesla.
And so on, until t = 10s.
By substituting various values of t into the expression for B, we can calculate the corresponding values of B. Plotting these values on a graph will give us the graph of B as a function of time.
b) The induced current, I(t), in the coil can be calculated using Faraday's law of electromagnetic induction. According to Faraday's law, the induced electromotive force (emf) and the rate of change of magnetic flux are related.
The induced electromotive force is given by the equation:
emf = -dϕ/dt
For a coil with N turns, the magnetic flux through the coil is given by:
ϕ = B * A
where B is the magnetic field strength and A is the area of the coil.
In this case, the coil has 20 turns, so the induced emf is:
emf = -N * dϕ/dt
= -20 * d/dt(B * A)
= -20A * dB/dt
The resistance of the coil is given as 0.50 Ω, so using Ohm's law, we can relate the induced emf to the induced current:
emf = I(t) * R
where I(t) is the induced current as a function of time, and R is the resistance of the coil.
By substituting the expression for the induced emf, we get:
-20A * dB/dt = I(t) * R
Now we need to find the expression for dB/dt, the rate of change of B with respect to time. We differentiate the expression for B with respect to t:
dB/dt = 0.020 + 0.020 = 0.040
Substituting the value of dB/dt into the previous equation, we get:
-20A * 0.040 = I(t) * R
-0.800A = I(t) * R
Therefore, the expression for the induced current, I(t), as a function of time is:
I(t) = -0.800A / R
c) To evaluate I at t = 5s and t = 10s, we need to substitute these values into the expression for I(t):
At t = 5s:
I(5) = -0.800A / R
At t = 10s:
I(10) = -0.800A / R
To fully evaluate I at t = 5s and t = 10s, we need to know the value of A, the area of the coil, and the resistance, R, of the coil.
(a) parabolic curve,
(b)
ε = -N•(dΦ/dt)= -N•(d(B•A•cosα)/dt),
cosα =1, A =π•d^2/4.
ε = I•R,
I•R = -N• A •(dB/dt),
I =- (N• π•d^2/4•R) •(0.01•2•t +0.02) =
= - (20• π•0.25•0.02/4•0.5)(t+1) =0.157(t+1),
(c)
t=5 s => I = 0.157•6 =0.942 A,
t= 10 s => I = 0.157•11=1.727 A.