An elliptical loop having resistance R, of semi major axis a and semi minor axis b is placed in a magnetic field as shown in the figure. If the loop is rotated about the x-axis with angular frequency ω, the average power loss in the loop due to joule heating is:

Option 1 -

π²a²b²B²ω² / R

Option 2 -

zero

Option 3 -

πabBω / R

Option 4 -

π²a²b²B²ω² / 2R

0 3 Views | Posted 2 months ago
Asked by Shiksha User

  • 1 Answer

  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    2 months ago
    Correct Option - 4


    Detailed Solution:

    For a rotating loop, the induced emf is ε = ε? sin (ωt), where the peak emf ε? = NABω.
    Here, Area A = πab. The number of turns N = 1.
    So, ε? = B (πab)ω
    The average power loss due to Joule heating in a resistor R is given by:
    P_avg = (ε_rms)² / R
    Where ε_rms = ε? /√2
    P_avg = (ε? ²/2) / R = (B (πab)ω)² / (2R) = π²a²b²B²ω² / (2R)

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