A rectangular loop of side 8cm and 2cm with a smal cut is movg out of uniform magnetic field of magnitude O.3t directed normal to d loop .wat is d emf developed acros d cut if d velocity of d loop is 1cm in a directed normal to d 1) longer side,2)short side of d loop ?for how loop does d induced voltage last in each case?
To calculate the emf (electromotive force) developed across the cut in the rectangle loop, we can use Faraday's Law of electromagnetic induction.
1) For the longer side of the loop:
The emf developed can be calculated using the formula:
ε = -N * d(Φ)/dt
where ε is the emf, N is the number of turns in the loop, and d(Φ)/dt is the rate of change of magnetic flux through the loop.
In this case, since the loop is moving out of a uniform magnetic field, the rate of change of magnetic flux can be calculated as the product of the magnetic field strength and the area of the loop.
Given:
Side of the loop (a) = 8 cm = 0.08 m
Side of the cut (b) = 2 cm = 0.02 m
Magnitude of the magnetic field (B) = 0.3 T
Velocity of the loop (v) = 1 cm = 0.01 m
Number of turns in the loop (N) = 1 (as it is not mentioned)
The area of the loop is given by A = a * b, so the area is 0.08 m * 0.02 m = 0.0016 m^2.
The rate of change of magnetic flux (d(Φ)/dt) is equal to B * A * v. Substituting the given values:
d(Φ)/dt = 0.3 T * 0.0016 m^2 * 0.01 m/s = 4.8 x 10^-6 Tm^2/s.
Since there is only one turn in the loop (N = 1), the emf (ε) across the cut is given by:
ε = - N * d(Φ)/dt = - (1) * 4.8 x 10^-6 Tm^2/s = -4.8 x 10^-6 V.
Therefore, the emf developed across the cut on the longer side of the loop is -4.8 x 10^-6 V.
2) For the shorter side of the loop:
Using the same approach as above, the area of the loop this time is A = b * a = 0.02 m * 0.08 m = 0.0016 m^2.
The rate of change of magnetic flux (d(Φ)/dt) can be calculated as before:
d(Φ)/dt = 0.3 T * 0.0016 m^2 * 0.01 m/s = 4.8 x 10^-6 Tm^2/s.
Again, since there is only one turn in the loop, the emf (ε) across the cut is given by:
ε = - N * d(Φ)/dt = - (1) * 4.8 x 10^-6 Tm^2/s = -4.8 x 10^-6 V.
So, the emf developed across the cut on the shorter side of the loop is also -4.8 x 10^-6 V.
For both cases, the induced voltage (emf) lasts as long as the loop remains in the magnetic field and continues to move.
I hope this explanation helps you understand how to calculate the induced emf and the duration of the induced voltage in each case.