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3 years ago

The electric field in a 3.0mm×3.0mm square aluminum wire is 1.0×10−2 V/m . What is the current in the wire?

1 answer:

4 0

So here is my answer. Given that the electric field in <span>a 3.0mm×3.0mm square aluminum wire is 1.0×10−2 V/m, this is how we find the current in the wire.
</span><span>First, we take the distance as 3mm or 0.003m. Using the formula E=V/d, where d is distance, v voltage and E electric field strength, we make V the subject, being V=Ed or 2.2*10^-2*0.003=6.6*10^-5V
</span>Hope this answers your question.

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A coil has 400 turns and self-inductance 7.50 mH. The current in the coil varies with time according to i = 1680 mA2 cos [πt/(0.

zhannawk [14.2K]

Answer:

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(b) 0.0126 Wb

Solution:

As per the question:

No. of turns in the coil, N = 400 turns

Self Inductance of the coil, L = 7.50 mH = $7.50\times 10^{- 3}\ H$

Current in the coil, i = $1680cos[\frac{\pi t}{0.0250}]$ A

where

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Now,

(a) To calculate the maximum emf:

We know that maximum emf induced in the coil is given by:

$e = \frac{Ldi}{dt}$

$e = L\frac{d}{dt}(1680)cos[\frac{\pi t}{0.0250}]$

$e = - 7.50\times 10^{- 3}\times \frac{\pi}{0.0250}\times \frac{d}{dt}(1680)sin[\frac{\pi t}{0.0250}]$

For maximum emf, $sin\theta$ should be maximum, i.e., 1

Now, the magnitude of the maximum emf is given by:

$|e| = 7.50\times 10^{- 3}\times 1680\times 10^{- 3}\times \frac{\pi}{0.0250} = 1.58\ V$

(b) To calculate the maximum average flux,we know that:

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