Question 4: EM waves

Question

The magnetic field of a beam emerging from a fitter facing a flood light is given by
B = 10 $$	imes$$ 10$$-$$8 sin(1 $$	imes$$ 107z $$-$$ 3.6 $$	imes$$ 1015t)T. The average intensity of the beam is

Accepted Answer

Given, magnetic field,
B = 10 $$	imes$$ 10
$$-$$8
sim(1 $$	imes$$ 10
7
z $$-$$ 3.6 $$	imes$$ 10
15
t) T ......... (i)
Standard equation of magnetic field,
B = B
0
sin(kz $$-$$ $$\omega$$t) ....... (ii)
On comparing Eqs. (i) and (ii), we get
B
0
= 10 $$	imes$$ 10
$$-$$8
T
The average intensity of the beam,
$${I_{av}} = {{B_0^2} \over {2{\mu _0}}} = {1 \over 2} 	imes {{{{(10 	imes {{10}^{ - 8}})}^2} 	imes 3 	imes {{10}^8}} \over {4\pi  	imes {{10}^{ - 7}}}}$$ [$$\because$$ $$c = 3 	imes {10^8}$$ ms
$$-$$1
]
= 1.19 W/m
2

Answer

1.82 W/m2

Answer

1.19 W/m2

Answer

1.18 W/m2

Answer

1.17 W/m2

EM waves

The magnetic field of a beam emerging from a fitter facing a flood light is given by B = 10 $\times 10-8 sin(1 \times 107z - 3.6 \times$ 1015t)T. The average intensity of the beam is

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