contestada

The magnitude of the magnetic flux through the surface of a circular plate is 5.90 10-5 T Ā· m2 when it is placed in a region of uniform magnetic field that is oriented at 42.0Ā° to the vertical. The radius of the plate is 8.50 cm. Determine the strength of the magnetic field.

Respuesta :

onyebuchinnaji

Answer:

The strength of the magnetic field is 3.5 x 10ā»Ā³ T

Explanation:

Given;

magnitude of the magnetic flux , Ī¦ = 5.90 x 10ā»āµ TĀ·mĀ²

angle of inclination of the field, Īø = 42.0Ā°

radius of the circular plate, r = 8.50 cm = 0.085 m

Generally magnetic flux in a uniform magnetic field is given as;

Ī¦ = BACosĪø

where;

B is the strength of the magnetic field

A is the area of the circular plate

Area of the circular plate:

A = Ļ€rĀ²

A = Ļ€ (0.085)Ā² = 0.0227 mĀ²

The strength of the magnetic field:

B = Ī¦ / ACosĪø

B = ( 5.90 x 10ā»āµ) / ( 0.0227 x Cos42)

B = 3.5 x 10ā»Ā³ T

Therefore, the strength of the magnetic field is 3.5 x 10ā»Ā³ T

The strength of the magnetic field should be 3.5 x 10ā»Ā³ T.

Calculation of the strength of the magnetic field:

Since

The magnitude of the magnetic flux, Ī¦ = 5.90 x 10ā»āµ TĀ·mĀ²

The angle of inclination of the field, Īø = 42.0Ā°

The radius of the circular plate, r = 8.50 cm = 0.085 m

Generally, the magnetic flux in a uniform magnetic field is provided as

Ī¦ = BACosĪø

Here,

B represents the strength of the magnetic field

A represents the area of the circular plate

Area of the circular plate:

So,

A = Ļ€rĀ²

A = Ļ€ (0.085)Ā² = 0.0227 mĀ²

Now

The strength of the magnetic field:

B = Ī¦ / ACosĪø

B = ( 5.90 x 10ā»āµ) / ( 0.0227 x Cos42)

B = 3.5 x 10ā»Ā³ T

Learn more about magnetic here: https://brainly.com/question/24968037