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

Answer :

onyebuchinnaji

Answer:

The strength of the magnetic field is 3.5 x 10⁻³ T

Explanation:

Given;

magnitude of the magnetic flux , Φ = 5.90 x 10⁻⁵ T·m²

angle of inclination of the field, θ = 42.0°

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

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

Φ = BACosθ

where;

B is the strength of the magnetic field

A is the area of the circular plate

Area of the circular plate:

A = πr²

A = π (0.085)² = 0.0227 m²

The strength of the magnetic field:

B = Φ / ACosθ

B = ( 5.90 x 10⁻⁵) / ( 0.0227 x Cos42)

B = 3.5 x 10⁻³ T

Therefore, the strength of the magnetic field is 3.5 x 10⁻³ T

The strength of the magnetic field should be 3.5 x 10⁻³ T.

Calculation of the strength of the magnetic field:

Since

The magnitude of the magnetic flux, Φ = 5.90 x 10⁻⁵ T·m²

The angle of inclination of the field, θ = 42.0°

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

Φ = BACosθ

Here,

B represents the strength of the magnetic field

A represents the area of the circular plate

Area of the circular plate:

So,

A = πr²

A = π (0.085)² = 0.0227 m²

Now

The strength of the magnetic field:

B = Φ / ACosθ

B = ( 5.90 x 10⁻⁵) / ( 0.0227 x Cos42)

B = 3.5 x 10⁻³ T

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