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Bezzdna [24]
2 years ago
6

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:_____

_.
Physics
1 answer:
Lady bird [3.3K]2 years ago
4 0

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.

To learn more about Solensid visit:

brainly.com/question/15061394

#SPJ4

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An investigator collects a sample of a radioactive isotope with an activity of 490,000 Bq.48 hours later, the activity is 110,00
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Answer:

The correct answer is "22.27 hours".

Explanation:

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As we know,

⇒ A = A_0 e^{- \lambda t}

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By taking "ln", we get

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⇒ \frac{ln_2}{T_ \frac{1}{2} } =0.031122

⇒ T_\frac{1}{2}=\frac{ln_2}{0.031122}

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