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a solenoid of length 1.50 cm and radius 0.650 cm has 49 turns. if the wire of the solenoid has 1.45 amps of current, what is the magnitude of the magnetic field inside the solenoid?

Respuesta :

The magnetic field of the solenoid will be 30.79 T.

Magnetic field strength is a quantitative measure of the strength or weakness of a magnetic field.

  • It is also the force experienced by a north pole unit of one weber force at a certain point in the magnetic field.
  • SI unit for the magnetic field is Tesla (T).
  • Magnetic field is denoted by B and has the formula as B = μ₀I/2πr
  • Also it has an another formula as  B = (N*r*I) /L

Length of solenoid, L = 1.5 cm = 0.015 m

Radius of solenoid, r = 0.65 cm = 0.0065 m

Number of turns, N = 49 turns

Current, I = 1.45 A

Magnetic field, B, can be calculated as B = (N*r*I) /L

B = (49 * 0.0065 * 1.45)/0.015

B =  30.79 T

The magnitude of the magnetic field inside the solenoid is 30.79 T.

To know more about magnetic field,

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