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JEE Main 2022 (24 Jun, Shift 2) · Energy Density & Intensity

Energy Density & Intensity question from JEE Main 2022 (24 Jun, Shift 2)

Q17JEE Main 2022 · 24 JUN · SHIFT 2Medium

Energy Density & Intensity

An electric bulb is rated as 200 W. What will be the peak magnetic field at 4 m distance produced by the radiations coming from this bulb? Consider this bulb as a point source with 3.5% efficiency.

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Answer: (b)

  1. Radiated power P=0.035×200=7 WP = 0.035 \times 200 = 7\ \text{W}
  2. I=P4πr2=74π×16≈0.0348 W/m2I = \dfrac{P}{4\pi r^2} = \dfrac{7}{4\pi \times 16} \approx 0.0348\ \text{W/m}^2
  3. I=cB022μ0⇒B0=2μ0Ic=2×4π×10−7×0.03483×108I = \dfrac{c B_0^2}{2\mu_0} \Rightarrow B_0 = \sqrt{\dfrac{2\mu_0 I}{c}} = \sqrt{\dfrac{2 \times 4\pi \times 10^{-7} \times 0.0348}{3 \times 10^8}}
  4. B0≈1.71×10−8 TB_0 \approx 1.71 \times 10^{-8}\ \text{T}, so the answer is (b).
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