Magnetic Field and induced current

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A square wire loop with a side length of 4.0 cm and a resistance of 2.0 x 10^(-3) ohms is placed in a vertical magnetic field of 0.7 T, which is switched off uniformly over 0.8 seconds. The induced electromotive force (EMF) can be calculated using the change in magnetic flux over time. The induced current can then be determined using Ohm's law, and the energy dissipated in the loop can be calculated using the induced current and the time duration. The discussion emphasizes the application of relevant formulas to solve for the induced current and energy dissipation. Understanding these calculations is essential for solving the problem effectively.
bchl85
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Can anyone help me to do this problem. Thanks in advance.

A closed wire loop in the form of a square of side 4.0cm is mounted with its plane horizontal. The loop has a resistance of 2.0*10^(-3) ohm. The loop is situated in the magnetic field of strength 0.7 T directed vertically downwards.
When the field is switched off, it decreases to 0 at a uniform rate in 0.8s
What is:
a) The current induced in the loop.?
b) The energy dissipated in the loop during the change in the magnetic field?
 
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What have you done so far? You have enough information to work out the EMF induced in the loop, so you can work out the current.
 
bchl85 said:
Can anyone help me to do this problem. Thanks in advance.

A closed wire loop in the form of a square of side 4.0cm is mounted with its plane horizontal. The loop has a resistance of 2.0*10^(-3) ohm. The loop is situated in the magnetic field of strength 0.7 T directed vertically downwards.
When the field is switched off, it decreases to 0 at a uniform rate in 0.8s
What is:
a) The current induced in the loop.?
b) The energy dissipated in the loop during the change in the magnetic field?

Here are some formulas that might help:
U_{induced}=-\frac{\Delta \Phi_{mag}}{\Delta t}
\Phi_{mag}=:\vec{B}\cdot \vec{S}
I_{induced}=\frac{U_{induced}}{R}
W=U_{induced}I_{induced}t

Daniel.

PS.The problem is very easy.Hopefully u'll solve it by yourself...
 
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