NEET-XII-Physics
38: Electromagnetic Induction
- #4A conducting circular loop having a radius of 5.0 cm, is placed perpendicular to a magnetic field of 0.50 T. It is removed from the field in 0.50 s. Find the average emf produced in the loop during this time.Ans : Given:
Magnetic field intensity, B = 0.50 T
Radius of the loop, r = 5.0 cm = 5 × 10-2 m
∴ Area of the loop, A = `` \,\mathrm{\,\pi \,}{r}^{2}``
Initial magnetic flux in the loop, ϕ1 = B × A
ϕ1 = 0.5 × `` \,\mathrm{\,\pi \,}``(5 × 10-2)2 = 125`` \,\mathrm{\,\pi \,}`` × 10-5
As the loop is removed from the magnetic field, magnetic flux (Ï•2) = 0.
Induced emf ε is given by
`` \,\mathrm{\,\epsilon \,}=\frac{{\,\mathrm{\,\varphi \,}}_{1}-{\,\mathrm{\,\varphi \,}}_{2}}{t}``
`` =\frac{125\,\mathrm{\,\pi \,}\times {10}^{-5}}{5\times {10}^{-1}}``
`` =25\,\mathrm{\,\pi \,}\times {10}^{-4}``
= 25 × 3.14 × 10-4
= 78.5 × 10-4 V = 7.8 × 10-3 V
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