Question

A 10-turn ideal solenoid has an inductance of 4. 0 mh. to generate an emf of 2. 0 v the current should change at a rate of:______.

Answer 1

The answer is 500 A/sec.

N=10 ; Solensid; L=4 m H ; $\varepsilon=2 \mathrm{~V};$

$\begin{aligned}&\varepsilon=-L \frac{d I}{d t}=N \phi ; \\&2 \times \frac{1}{L}=-\frac{d I}{d t} ; \\&2 \times \frac{1}{4 \times 10^{-3}}=-\frac{d I}{d t} ;\end{aligned}$

$0.5 \times 10^{3}=\frac{d I}{dt}$

So, 500 A/sec.

What is Solensid?

• Electromagnets are magnets in which the wire is coiled around an iron core. When an electric current is applied to the iron core, a magnetic field is formed around it. When the power is turned off, the magnetic field dissipates. The wire-wound core of the magnet is ferromagnetic or ferrimagnetic in nature. The most prevalent material utilised in the core is iron.
• The magnetic field created is governed by the availability of power. A permanent magnet does not require electricity, and its magnetic field cannot be changed. Solenoid, MRI machines, hard drives, relays, motors, loudspeakers, and generators all employ electromagnets.

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