Question

VP 29.11.1 A long solenoid has a cross-sectional area of 3.00 cm2cm2. The current through the windings is decreasing at a rate of 22.0 A/sA/s. A wire loop of radius 3.10 cmcm is around the solenoid, parallel with its coils, and centered on the axis of the solenoid. The magnitude of the induced emf is 15.0 μVμV. Find the number of turns per meter in the solenoid.

Answer 1

To find the number of turns per meter in the solenoid, we can use Faraday's law of electromagnetic induction.

Step-by-step explanation:

To find the number of turns per meter in the solenoid, we can use Faraday's law of electromagnetic induction.

The induced emf in the wire loop is given by the equation:

emf = -N(dΦ/dt)

where N is the number of turns in the solenoid, Φ is the magnetic flux, and dt is the change in time.

We are given that the magnitude of the induced emf is 15.0 μV. Using the equation above and the given data, we can solve for N.

Answer 2

The number of turns per meter in the solenoid is 42 turns/meter.

How to calculate the number of turns?

The number of turns per meter in the solenoid is calculated as follows;

emf = NBA/t

where;

• B is the magnetic field strength
• N is the number of turns
• A is the area of the coil
• t is the time

The area of the coil is calculated as follows;

A = 3 cm²

A = 0.0003 m²

The magnetic field strength is;

B = μNI/L

B = (4π x 10⁻⁷ x N x 22) / L

B = 2.77 x 10⁻⁵ TN/Ls

emf = N(2.77 x 10⁻⁵ N) A

15 x 10⁻⁶ = N² x 2.77 x 10⁻⁵ x 0.0003

N² = 1,808.58

N = √ (1,808.58)

N = 42 turns/meter