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The intensity of the light from a bulb incident on a surface is 0.22 W/m^{2} . The amplitude of the magnetic field in this light-wave is ______× 10^{–9 }T.

(**Given: **Permittivity of vacuum ε_{0} = 8.85 × 10^{–12} C^{2 }N^{–1} – m^{–2}, speed of light in vacuum c = 3 × 10^{8} ms^{-1})

#### Options

46 × 10

^{–9}T42 × 10

^{–9}T45 × 10

^{–9}T43 × 10

^{–9}T

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#### Solution

The intensity of the light from a bulb incident on a surface is 0.22 W/m^{2} . The amplitude of the magnetic field in this light-wave is **43** × 10^{–9 }T.

**Explanation:**

As I = `(1/2 epsilon_0"E"_0^2)`

On substituting value of I, ε_{0} and c, we get

0.22 = `1/2 (8.85 xx 10^-12) "E"_0^2xx3xx10^8`

E_{0 }= `sqrt((2xx0.22)/(8.85xx10^-12xx3xx10^8))`

= 12.9 N/C

As c = `"E"_0/"B"_0`

⇒ B_{0} = `12.9/(3xx10^8)`

= 4.3 × 10^{–8 }

= 43 × 10^{–9} T

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