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3 years ago

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What current is required in the windings of a long solenoid that has 420 turns uniformly distributed over a length of 0.575 m in

order to produce inside the solenoid a magnetic field of magnitude 4.22×10−5 T? The permeablity of free space is 1.25664 × 10 Tm/A.

1 answer:

WARRIOR [948]3 years ago

4 0

Answer:

The current in the solenoid=0.463 Ampere

Explanation:

Given:

Number of turns , N=420

Length of the solenoid=0.575 m

Magnitude of Magnetic Field, $B=4.22\times10^{-5}\ \rm T$

We know that the magnitude of the magnetic Field inside the solenoid is

$B=\mu_0 ni$

Where

$\mu_0$ is the permeability of free space.
n is the number of turns per unit length
i is the current

According to question

$B=\mu_0 ni\\\\4.22\times10^{-5}=1.24664\times10^{-7}\times \dfrac{420}{0.575}\times i\\=0.463\ \rm Amp$

Hence the current is calculated.

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Answer:

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I hope this is useful for you

regards

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3 years ago