A loop wire is in a magnetic field of 2T. What is the magnitude of the magnetic flux?
0.2 m radius
B = 2T
30 degrees
and n is pointing out of the page
Ok so BAcos(30)
B= 2T
A= 3.14 r^2 for a circle right? (3.14)(0.2)^2
cos(30)
2T(3.14)(0.2)^2(cos(30))= 0.0335
and the answer is .22Wb??
Yes, if you have the orientation correctly.
what do you mean by orientation ?
never mind i got it thanks!
To find the magnitude of the magnetic flux, we need to multiply the magnetic field (B) by the area (A) of the loop and the cosine of the angle (θ) between the magnetic field and the normal to the loop.
In this case, we have:
B = 2T (given)
A = πr^2 = π(0.2m)^2 = 0.1256 m^2 (radius, r = 0.2m)
θ = 30 degrees
Now, let's plug in these values into the formula:
Flux (Φ) = B * A * cos(θ)
= 2T * 0.1256 m^2 * cos(30)
= 0.2512 T m^2 * cos(30)
Using the value of cos(30) = √3/2, we have:
Flux (Φ) = 0.2512 T m^2 * (√3/2)
≈ 0.2177 T m^2
Therefore, the magnitude of the magnetic flux is approximately 0.2177 T m^2.