Worksheet: Self-Inductance of a Long Solenoid

In this worksheet, we will practice calculating the self-inductance of a cylindrical solenoid and the emf induced by it when the current through it varies with time.

Q1:

A solenoid 80.0 cm long is wound with 600 turns of wire. The cross-sectional area of the coil is 6.00 cm2. What is the self-inductance of the solenoid?

  • A 3 . 4 0 × 1 0 H
  • B 7 . 6 1 × 1 0 H
  • C 9 . 9 2 × 1 0 H
  • D 5 . 6 7 × 1 0 H
  • E 6 0 . 7 × 1 0 H

Q2:

A long, cylindrical solenoid with 100 turns per centimeter has a radius of 1.5 cm.

Neglecting end effects, what is the self-inductance per unit length of the solenoid?

If the current through the solenoid changes at the rate 5.0 A/s, what is the emf induced per unit length?

Q3:

A cylindrical solenoid is wrapped around an iron core. The self-inductance of the solenoid is given by 𝐿=(1+𝑥)𝜇𝑁𝐴𝑙, where 𝑙 is the solenoid’s length, 𝐴 its cross-sectional area, and 𝑁 is the number of its turns. The solenoid’s core has a magnetic susceptibility 𝑥 of 4.0×10 and the current in the solenoid is 2.0 A.

What is the strength of the magnetic field in the core?

What is the effective surface current formed by the atomic current loops in the core?

What is the self-inductance of the solenoid?

Q4:

A rectangular toroid with inner radius 𝑅=8.0cm, outer radius 𝑅=11.0cm, height =5.0cm, and 𝑁=2,500 turns is filled with an iron core of magnetic susceptibility 5.2×10.

What is the self-inductance of the toroid?

If the current through the toroid is 0.5 A, what is the magnetic field at the center of the core?

Q5:

Consider a solenoid that has a core made of chromium. The solenoid has a length of 1.50 m, a cross-sectional area of 60.0 cm2, and a total of 50 turns. Chromium has a magnetic susceptibility of 3.10×10. Calculate the self-inductance of the solenoid.

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