What is the magnitude of the magnetic field at the point the center of a long, straight solenoid; the solenoid carries current 0.300 A and has 4.00 * 103 windings per meter

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29-07-2022
Physics

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What is the magnitude of the magnetic field at the point the center of a long, straight solenoid; the solenoid carries current 0.300 A and has 4.00 * 103 windings per meter

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rahulsharma789888 rahulsharma789888 · Answer 1 · 2022-08-12T07:54:32+02:00

The magnitude of the magnetic field at the point at the center of a long, straight solenoid if the solenoid carries a current of 0.300 A and has 4.00 ×10³ windings per meter would be 1.512 ×10⁻³ N/Amps m

What is a magnetic field?

A magnetic field could be understood as an area around a magnet, magnetic material, or an electric charge in which magnetic force is exerted. The SI unit of the magnetic field is tesla.

The magnetic field at the point at the center of a long, straight solenoid can be calculated by using the formula

B = μnI

where B is the magnetic field

μ is the Permeability constant having a value of 1.26 × 10⁻⁶ T/m

n is the number of winding per meter

I represent the current

By substituting the respective value in the formula

B=( 1.26 × 10⁻⁶ )× (4.00 ×10³)×0.300

B =1.512 ×10⁻³ N/Amps m

Thus, the magnitude of the magnetic field at the point at the center of a long, straight solenoid would be 1.512 ×10⁻³ N/Amps m.

Learn more about the magnetic fields from here

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