A loop of area 0.250 m^2 is in a uniform 0.020 0-T magnetic field. If the flux through the loop is 3.83 × 10-3T· m2, what angle does the normal to the plane of the loop make with the direction of the magnetic field?

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abidemiokin abidemiokin · Answer 1 · 2020-10-21T18:06:32+02:00

Answer:

40.0⁰

Explanation:

The formula for calculating the magnetic flux is expressed as:

[tex]\phi = BAcos\theta[/tex] where:

[tex]\phi[/tex] is the magnetic flux

B is the magnetic field

A is the cross sectional area

[tex]\theta[/tex] is the angle that the normal to the plane of the loop make with the direction of the magnetic field.

Given

A = 0.250m²

B = 0.020T

[tex]\phi[/tex] = 3.83 × 10⁻³T· m²

3.83 × 10⁻³ = 0.020*0.250cosθ

3.83 × 10⁻³ = 0.005cosθ

cosθ = 0.00383/0.005

cosθ = 0.766

θ = cos⁻¹0.766

θ = 40.0⁰

Hence the angle normal to the plane of the loop make with the direction of the magnetic field is 40.0⁰