Problem 25.49
A 1.40-m-diameter MRI solenoid with a length of 2.80 m has a magnetic field of 2.50 T along its axis.
Part A -
If the current is turned off in a time of 1.2 s, what is the emf in one turn of the solenoid's windings?
Express your answer using two significant figures
To find the emf in one turn of the solenoid's windings, we can use Faraday's law of electromagnetic induction, which states that the emf induced in a circuit is equal to the rate of change of magnetic flux through the circuit.
The magnetic flux, Φ, through a solenoid is given by the formula:
Φ = B * A
Where B is the magnetic field along the axis of the solenoid and A is the area of the cross-section of the solenoid.
Given:
Diameter of the solenoid (d) = 1.40 m
Radius of the solenoid (r) = d/2 = 1.40 m/2 = 0.70 m
Length of the solenoid (L) = 2.80 m
Magnetic field along the axis (B) = 2.50 T
To find the area of the cross-section of the solenoid, we can use the formula for the area of a circle:
A = π * r^2
Substituting the values:
A = π * (0.70 m)^2 = 1.54 m^2
Now, we can calculate the magnetic flux through the solenoid:
Φ = B * A = 2.50 T * 1.54 m^2 = 3.85 Wb
Since the current is turned off in a time of 1.2 s, the rate of change of magnetic flux, dΦ/dt, is equal to the magnetic flux divided by the time:
dΦ/dt = Φ / t = 3.85 Wb / 1.2 s = 3.21 V
Therefore, the emf in one turn of the solenoid's windings is 3.21 V.
To find the emf (electromotive force) in one turn of the solenoid's windings, we need to use Faraday's law of electromagnetic induction. This law states that the emf induced in a circuit is equal to the rate of change of magnetic flux through the loop.
The formula for calculating the emf using Faraday's law is:
emf = -N * (ΔΦ/Δt)
where:
- emf is the electromotive force
- N is the number of turns in the solenoid's windings
- ΔΦ is the change in magnetic flux
- Δt is the change in time
In this case, we are given the diameter of the solenoid (1.40 m), which means the radius (r) is half of the diameter (r = 0.70 m). The length of the solenoid is 2.80 m, and the magnetic field (B) along its axis is 2.50 T.
The magnetic flux (Φ) through a solenoid is given by the formula:
Φ = B * A
where:
- Φ is the magnetic flux
- B is the magnetic field
- A is the cross-sectional area of the solenoid
The cross-sectional area of a solenoid can be calculated using the formula:
A = π * r^2
Substituting the given values into the formulas, we get:
A = π * (0.70 m)^2
A ≈ 1.54 m^2
ΔΦ = B * A
ΔΦ = 2.50 T * 1.54 m^2
ΔΦ ≈ 3.85 T m^2
Now, we can calculate the emf:
emf = -N * (ΔΦ/Δt)
emf = -1 * (3.85 T m^2 / 1.2 s)
emf ≈ -3.21 V
Since the question asks for the answer to be expressed using two significant figures, we can round the emf to -3.2 V.