Question

A solenoid with 465 turns has a length of 8.00 cm and a cross-sectional area of 3.10 ✕ 10−9 m2. Find the solenoid's inductance and the average emf around the solenoid if the current changes from +3.50 A to −3.50 A in 7.83 ✕ 10−3 s

Answer 1

The solenoid's inductance and the average emf around the solenoid if the current changes from +3.50 A to −3.50 A in 7.83 ✕ 10−3 s will be 1.01 ×10⁻⁸ H and 9.02 × 10⁻² V.

What is a solenoid?

A coil of wire that carries an electric current is a solenoid. A solenoid is an electromagnet formed by a helical coil of wire.

Which generates a magnetic field when an electric current is passed through the coil.

A solenoid is a form of coil that produces a magnetic field when the electric current is passed through it. A solenoid is created when a conductive wire is used to make a loop.

Given data;

Turns ,N = 465

Length,L= 8.00 cm

Cross-sectional area,A = 3.10 ✕ 10−9 m2.

Solenoid's inductance, L=?

The average emf around the solenoid, E=?

Time of flow,t= 7.83 ✕ 10−3 s

Current changes from +3.50 A to −3.50 A

The inductance of the solenoid is found as;

$\rm L = \frac{\mu_0 AN^2}{L} \\\\ \rm L = \frac{4 \times \pi \times 10^{-7}\times 3.0 \times 10^{-9} \times (465)^2 }{8.00 \times 10^{-2}} \\\\ L= 1.01 \times 10^{-8} \ H$

The average emf around the solenoid is found as;

$\rm e = L \frac{I_2-I_1}{t} \\\\ \rm e = 1.01 \times 10^{-8} \times \frac{3.50-(-3.50)}{7.83 \times 10^{-3}} \\\\ e =9.02 \times 10^{-12} \ V$

Hence, the solenoid's inductance will be 1.01× 10⁻⁸ H.

To learn more about the solenoid refer;

brainly.com/question/16015159

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