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A wire 40 cm long bent into a rectangula...

A wire 40 cm long bent into a rectangular loop `15 cmxx5 cm` is placed perpendicular to the magnetic field whose flux density is `0.8 Wbm ^(-2)`. Within 1.0 second, the loop is changed into a 10 cm square and flux density increases to 1.4 `Wbm^(-2)`. Calculate the value of induced emf.

Text Solution

Verified by Experts

The correct Answer is:
`0.02`
Initial area of the loop,
`A = (15cm xx 5cm) = 75 cm^2 = 75xx10^(-4)m^2`
Initial flux density, `B = 0.8Wb//m^2`
Initial flux, `phi_B = BA = (0.8Wb//m^2)(75xx10^(-4)m^2) = 0.006Wb `
Final area of the loop, A. `= (10cm xx 10 cm) = 100 cm^2`
`= 10^(-2)m^2`
Final flux density, `B.=1.4Wb//m^2`
Final flux, `phi_B` = B.A. `= (1.47Wb//m^(2))(10^(-2)m^2)`
= 0.014 Wb
Change in flux,
`phi._(B)-phi_(B) = (0.014 -0.006) Wb=0.008 Wb`
Time during which this change in flux takes place, t = 0.5 s
Induced emf, `E =(phi-_B=phi_B)/(t)=(0.008)/(0.5) = 0.016 ~~ 0.02V `
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