Questions: A long solenoid of radius r=2.00 cm is wound with 350 turns / m and carries a current that changes at the rate of 28.5 A / s as in Figure P20.60. What is the magnitude of the emf induced in the square conducting loop surrounding the center of the solenoid?

Solution Steps

Given:

• Radius of solenoid, \( r = 2.00 \, \text{cm} = 0.02 \, \text{m} \)
• Number of turns per meter, \( n = 3.50 \times 10^3 \, \text{turns/m} \)
• Rate of change of current, \( \frac{dI}{dt} = 28.5 \, \text{A/s} \)

We need to find the magnitude of the emf induced in the square conducting loop surrounding the center of the solenoid.

Relevant formula: \[ \text{emf} = -\frac{d\Phi_B}{dt} \] where \( \Phi_B \) is the magnetic flux.

The magnetic field inside a solenoid is given by: \[ B = \mu_0 n I \] where \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \) is the permeability of free space.

The magnetic flux \( \Phi_B \) through the loop is: \[ \Phi_B = B \cdot A \] where \( A \) is the area of the loop. Since the loop surrounds the solenoid, the area \( A \) is: \[ A = \pi r^2 \]

The rate of change of magnetic flux is: \[ \frac{d\Phi_B}{dt} = \frac{d(B \cdot A)}{dt} = A \cdot \frac{dB}{dt} \]

Since \( B = \mu_0 n I \), we have: \[ \frac{dB}{dt} = \mu_0 n \frac{dI}{dt} \]

Substitute \( \frac{dB}{dt} \) into the flux equation: \[ \frac{d\Phi_B}{dt} = A \cdot \mu_0 n \frac{dI}{dt} \]

Now, substitute the values: \[ A = \pi (0.02)^2 = 1.25664 \times 10^{-3} \, \text{m}^2 \] \[ \frac{dB}{dt} = (4\pi \times 10^{-7}) (3.50 \times 10^3) (28.5) \]

Calculate \( \frac{dB}{dt} \): \[ \frac{dB}{dt} = 4\pi \times 10^{-7} \times 3.50 \times 10^3 \times 28.5 \] \[ \frac{dB}{dt} \approx 1.25 \times 10^{-1} \, \text{T/s} \]

Finally, calculate the emf: \[ \text{emf} = A \cdot \frac{dB}{dt} \] \[ \text{emf} = 1.25664 \times 10^{-3} \times 1.25 \times 10^{-1} \] \[ \text{emf} \approx 1.57 \times 10^{-4} \, \text{V} \]

Final Answer