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statuscvo [17]
3 years ago
10

Will an electromagnet be obtained if steel is placed inside the solenoid?​

Physics
1 answer:
iogann1982 [59]3 years ago
7 0

Answer:A solenoid is a simple electromagnetic device consisting of a coiled electric wire, wrapped in a 3D circular pattern. When electric current is passed through the wire, the solenoid acts like a magnet with N and S poles at the ends of the coil.

When a ferromagnetic material rod is permanently placed inside the solenoid, the metal greatly increases the magnetic effect and becomes a permanent electromagnet. Moreover, it can also be used as an electrical switch by drawing in or pushing out a ferromagnetic material like an iron rod. Depending on the directions of the rod and the electrical current the switching action takes place.

Given figure  represents the solenoid as electromagnet and the switching action.

Explanation:

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\frac{U_B}{U_E} = \frac{1}{2\mu_o} B^2 / \frac{1}{2}\epsilon E^2\\\\\frac{U_B}{U_E} = \frac{B^2}{2\mu_o}  *\frac{2}{\epsilon E^2} \\\\\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{B^2}{E^2})

\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{B}{E})^2

But, Electric field potential, V = E x L = IR (I is current and R is resistance)

\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{B*L}{E*L})^2

Now replace E x L with IR

\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{B*L}{IR})^2

Also, B = μ₀I / 2πr, substitute this value in the above equation

\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{\mu_oI*L}{2\pi r* IR})^2

cancel out the current "I" and factor out μ₀

\frac{U_B}{U_E} = \frac{\mu_o^2}{\mu_o \epsilon} (\frac{L}{2\pi r* R})^2

Finally, the equation becomes;

\frac{U_B}{U_E} = \frac{\mu_o}{\epsilon} (\frac{L}{2\pi r*R })^2

Therefore, the correct option is (d) μ₀/ϵ₀ (L /R 2πr)²

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