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# Physics magnetic flux problem. Please Help!!! Will award best answer!!?

A magnetic field has a magnitude of B=.95T. It is directed perpendicular to a circular loop of wire that has a radius of .15m. The loop of wire rotated 90 degrees so that the magnetic field is parallel to the loop. What is the change in magnetic flux through the loop?

### 3 Answers

- VamanLv 73 years ago
0. lux = field strength* area. The area exposed to the field strength is 0. Therefore, the magnetic field strength is 0.

- dogsafireLv 73 years ago
Wouldn't the change in magnetic flux be 0.95T? The coil was orthogonal to the direction of the flux and now it's parallel. There is no magnetic flux _through_ the loop

- Anonymous3 years ago
phi = int_0^(2πθ) int_0^(0.15) 0.95*r dr dθ = 0.021375πθ;

dphi/dt = dphi/dθ * dθ/dt = emf = 0.021375πθ;

dphi/dt = 0.021375π*(dθ/dt) is the final answer because we do not know what the rate of the change of the loop was but that it was

rotating angular.

If you use phi = N*B*A*cos(theta);

dphi/dt = N*B*A*(-sin(θ))*(dθ/dt) = (1)*(0.95)*((0.15)^2)*π*(-sin(90 degrees))*(dθ/dt) = -0.021375π*(dθ/dt)

My guess is the answer is zero because θ = π/2; dθ/dt = 0;