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seraphim [82]
3 years ago
5

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.
Physics
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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