A uniform magnetic field passes through a horizontal circular wire loop at an angle 15.1° from the normal to the plane of the loop. The magnitude of the magnetic field is 3.35 T , and the radius of the wire loop is 0.240 m . Find the magnetic flux Φ through the loop.

Answers

Answer 1

Answer:

0.5849Weber

Explanation:

The formula for calculating the magnetic flus is expressed as:

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

Given

The magnitude of the magnetic field B = 3.35T

Area of the loop = πr² = 3.14(0.24)² = 0.180864m²

angle of the wire loop θ = 15.1°

Substitute the given values into the formula:

[tex]\phi = 3.35(0.180864)cos15.1^0\\\phi =0.6058944cos15.1^0\\\phi =0.6058944(0.9655)\\\phi = 0.5849Wb[/tex]

Hence the magnetic flux Φ through the loop is 0.5849Weber


Related Questions

While standing at the edge of the roof of a building, a man throws a stone upward with an initial speed of 6.63 m/s. The stone subsequently falls to the ground, which is 14.5 m below the point where the stone leaves his hand.

At what speed does the stone impact the ground? Ignore air resistance and use =9.81 m/s2 for the acceleration due to gravity.
How much time is the stone in the air?
elapsed time:

Answers

Answer:

Speed=28.1m/s(to 3s.f.) , Time=2.19s(to 3s.f.)

Explanation:

Time=Distance/Speed

=14.5/6.63

=2.19s(to 3s.f.)

Acceleration=Final Velocity(v)-Initial Velocity(u)/Time

9.81=v-6.63/2.19

v-6.63=21.5

v=28.1m/s

why is it necessary to have end correction in the organ pipe?​

Answers

Answer:

The vibrating length of the air column is greater than the actual length of the organ pipe

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