Faraday's Law of Induction and wire loop

[AznBoi]

Homework Statement

A wire loop of radius 0.3m lies so that an external magnetic field of magnitude 0.30 T is perpendicular to the loop. The field reverses its direction, and its magnitude changes to 0.20 T in 1.5s. Find the magnitude of the average induced emf in the loop during this time.

Homework Equations

$$\varepsilon_{avg} = \frac{-\Delta \Phi_{B}}{\Delta t}$$

$$\Phi_{B}=BAcos \theta$$

The Attempt at a Solution

$$A=\pi r^{2}=\pi (0.30m)^{2}= 0.09\pim^{2}$$

$$0s: \Phi_{B}=BA cos \theta =(0.30T)(0.09\pi m^{2}) = 0.0848 T*m^{2}$$

$$1.5s: \Phi = -(0.20T)(0.09\pi m^{2})= -0.05655 T*m^{2}$$

$$\varepsilon_{avg}=\frac{(-0.05655-0.0848)}{1.5s}=-0.14135$$

The correct answer, however, is 94 mV??

 

[AznBoi]

Nvm, I found my mistakes. Just some arithmetic errors :grumpy:

 

[m2003]

I would like to point out that the equation used in the attempt at a solution is incorrect. The correct equation for Faraday's Law of Induction is \varepsilon = -N\frac{\Delta \Phi_B}{\Delta t}, where N is the number of turns in the wire loop. This equation takes into account the number of turns in the loop, which is a crucial factor in determining the induced emf.

Using the correct equation, we can calculate the induced emf as follows:

N=1 (as there is only one turn in the wire loop)

0s: \Phi_B = BA cos \theta = (0.30T)(0.09\pi m^2) = 0.0848 T*m^2

1.5s: \Phi_B = -(0.20T)(0.09\pi m^2) = -0.05655 T*m^2

\Delta \Phi_B = (-0.05655 T*m^2 - 0.0848 T*m^2) = -0.14135 T*m^2

\Delta t = 1.5s - 0s = 1.5s

\varepsilon_{avg} = -N\frac{\Delta \Phi_B}{\Delta t} = -(1)(\frac{-0.14135 T*m^2}{1.5s}) = 0.094 mV

Therefore, the magnitude of the average induced emf in the loop during this time is 94 mV. It is important to use the correct equation in order to obtain accurate results in scientific calculations.