A wire of 50cm long bent in to a rectangular loop of 16cm by 4cm is placed perpendicular to the magnetic field whose flux density is 800Wb/m^2 with in 2 second.and the loop is changed in to 10cm square &the flux density increase to 1600Wb/m^2. Compute indused electromotive force ????
Given that the wire of 50 cm long is bent into a rectangular loop of 16 cm by 4 cm, we can calculate the area (A1) of this loop using the formula:
A1 = length * width
A1 = 16 cm * 4 cm = 64 cm^2 = 0.0064 m^2 (convert the area to square meters)
The magnetic field density (B1) is given as 800 Wb/m^2.
Now, the loop is changed into a 10 cm square. The area (A2) of this square can be found using the formula:
A2 = length²
A2 = 10 cm * 10 cm = 100 cm^2 = 0.01 m^2 (convert the area to square meters)
The flux density (B2) is given as 1600 Wb/m^2.
To find the induced electromotive force, we first need to find the change in magnetic flux Φ. The magnetic flux Φ is the product of the magnetic field density and the area of the loop:
Φ1 = A1 * B1 = 0.0064 m^2 * 800 Wb/m^2 = 5.12 Wb
Φ2 = A2 * B2 = 0.01 m^2 * 1600 Wb/m^2 = 16 Wb
Now, we can find the change in magnetic flux (ΔΦ):
ΔΦ = Φ2 - Φ1 = 16 Wb - 5.12 Wb = 10.88 Wb
We are given the time (t) of 2 seconds.
The induced electromotive force can be found using Faraday's law:
ε = -ΔΦ/t
ε = -(-10.88 Wb / 2 s) = 5.44 V
Therefore, the induced electromotive force is 5.44 V.