A coil with 210 turns, a radius of 5.2 cm, and a resistance of 11 ohms surrounds a solenoid with 160 turns/cm and a radius of 4.6 cm . The current in the solenoid changes at a constant rate from 0 to 6.4 A in 0.11 s.
Calculate the magnitude of the induced current in the coil.
To calculate the magnitude of the induced current in the coil, we can use Faraday's law of electromagnetic induction. According to Faraday's law, the magnitude of the induced electromotive force (EMF) in a closed loop is equal to the rate of change of magnetic flux through the loop.
The magnetic flux through a loop is given by the formula: Φ = B * A, where B is the magnetic field and A is the area of the loop.
To calculate the induced current, we need to calculate the magnitude of the induced EMF and then divide it by the resistance of the coil.
First, let's calculate the magnetic field produced by the solenoid. The formula for the magnetic field within a solenoid is: B = μ₀ * n * I, where μ₀ is the permeability of free space (4π * 10^-7 T m/A), n is the number of turns per unit length, and I is the current in the solenoid.
Using the given values, we have:
B = (4π * 10^-7 T m/A) * (160 turns/cm) * (6.4 A)
B = 4.096 × 10^-3 T
Next, let's calculate the area of the coil. The area of a coil is given by the formula: A = π * r^2, where r is the radius of the coil.
Using the given values, we have:
A = π * (5.2 cm)^2
A = 84.948 cm^2
Now, let's calculate the rate of change of magnetic flux through the coil. The rate of change of magnetic flux is given by the formula: dΦ/dt = B * dA/dt, where dΦ/dt is the rate of change of magnetic flux and dA/dt is the rate of change of area.
Since the radius of the solenoid remains constant, the rate of change of area is 0. Therefore, dA/dt = 0.
Now, let's calculate the rate of change of magnetic flux using the formula:
dΦ/dt = B * dA/dt
dΦ/dt = (4.096 × 10^-3 T) * (0 cm^2/s)
dΦ/dt = 0 T m^2/s
Finally, let's calculate the induced EMF using Faraday's law:
EMF = -dΦ/dt
EMF = -0 T m^2/s
Since the induced EMF is zero, this implies that there is no induced current in the coil.
Therefore, the magnitude of the induced current in the coil is 0 A.