To solve this problem, we need to use Faraday's law of electromagnetic induction, which states that the induced electromotive force (emf) in a loop is equal to the rate of change of magnetic flux through that loop. In this case, the rectangular loop is moving out of a magnetic field, which creates a change in flux and thus induces an emf across the cut. Let's break down the problem step by step for both scenarios: when the loop moves in the direction of the longer side and when it moves in the direction of the shorter side.
Understanding the Geometry
The rectangular wire loop has dimensions of 8 cm (longer side) and 2 cm (shorter side), and it is moving out of a uniform magnetic field of strength 0.3 T. The area of the loop is:
- Area (A) = Length × Width = 8 cm × 2 cm = 16 cm²
Calculating the Induced EMF
According to Faraday's law, the induced emf (ε) can be expressed by the formula:
ε = -dΦ/dt
Where Φ is the magnetic flux given by the equation:
Φ = B × A
As the loop moves out of the magnetic field, the effective area exposed to the magnetic field changes, which in turn changes the magnetic flux. The induced emf will depend on how quickly this area is being reduced.
a) Moving Normal to the Longer Side (8 cm)
When the loop is moving out in the direction normal to the longer side, the width of the loop that is moving out of the magnetic field is the shorter side (2 cm). The rate of change of area with respect to time can be calculated as follows:
- Width of the cut area moving out = 2 cm = 0.02 m
- Velocity (v) = 1 cm/s = 0.01 m/s
The change in area per unit time (dA/dt) is simply the velocity multiplied by the width:
dA/dt = v × width = 0.01 m/s × 0.02 m = 0.0002 m²/s
Now we can find the induced emf:
ε = B × (dA/dt) = 0.3 T × 0.0002 m²/s = 0.00006 V = 60 µV
b) Moving Normal to the Shorter Side (2 cm)
In this case, when the loop moves in the direction normal to the shorter side, the width of the loop that is moving out of the magnetic field is the longer side (8 cm). Again, we can calculate the change in area per unit time:
- Width of the cut area moving out = 8 cm = 0.08 m
Now, the change in area per unit time (dA/dt) becomes:
dA/dt = v × width = 0.01 m/s × 0.08 m = 0.0008 m²/s
Substituting this into the emf formula gives us:
ε = B × (dA/dt) = 0.3 T × 0.0008 m²/s = 0.00024 V = 240 µV
Duration of Induced Voltage
The duration of the induced voltage can be calculated by determining how long it takes for the entire loop to exit the magnetic field. The time (t) is given by the distance over velocity:
- For the longer side: Distance = 8 cm = 0.08 m, Velocity = 0.01 m/s
- Time (t) = Distance / Velocity = 0.08 m / 0.01 m/s = 8 s
- For the shorter side: Distance = 2 cm = 0.02 m, Velocity = 0.01 m/s
- Time (t) = Distance / Velocity = 0.02 m / 0.01 m/s = 2 s
Summary of Results
To summarize, the induced emf and duration for each case are:
- Moving normal to the longer side: Induced emf = 60 µV, Duration = 8 s
- Moving normal to the shorter side: Induced emf = 240 µV, Duration = 2 s
This analysis illustrates how the orientation and motion of the loop in the magnetic field directly affect the induced emf and the duration of that emf. The larger the area changing with respect to time, the greater the induced voltage. If you have any further questions on this topic, feel free to ask!
