Question

a circular loop of area 0.05m2 rotates in a uniform magnetic field of 0.2 tesla. If the loop rotates about its diameter which is perpendicular to the field find the flux linked with the loop when its plane is inclined 60 degree to the field?

Answer 1

The magnetic flux linked with the circular loop when its plane is inclined at 60 degrees to the magnetic field is approximately 0.01 Weber.

Step-by-step explanation:

To find the magnetic flux linked with the circular loop when its plane is inclined at 60 degrees to the magnetic field, we can use the formula for magnetic flux:

Φ = B * A * cos(θ)

where:

Φ = magnetic flux

B = magnetic field strength (in Tesla)

A = area of the loop (in square meters)

θ = angle between the magnetic field and the plane of the loop (in radians)

Given:

B = 0.2 Tesla

A = 0.05 m^2

θ = 60 degrees = 60 * π / 180 radians (convert to radians)

Let's calculate the magnetic flux:

θ = 60 * π / 180

= π / 3 radians

Φ = 0.2 Tesla * 0.05 m^2 * cos(π / 3)

≈ 0.2 * 0.05 * 0.5 (cos(π / 3) ≈ 0.5)

≈ 0.01 Weber

So, the magnetic flux linked with the circular loop when its plane is inclined at 60 degrees to the magnetic field is approximately 0.01 Weber.