$\begin {array} {1 1} (a)\;0.26\: m \: and \: 239 \: MHz & \quad (b)\;1.26\: cm \: and \: 23.9 \: GHz \\ (c)\;1.26\: m \: and \: 23.9 \: MHz & \quad (d)\;0.26\: cm \: and \: 2.39 \: GHz \end {array}$

Comparing the given equation with $B_y = B_o \: \sin \bigg[2 \pi \bigg( \large\frac{x}{\lambda} + \large\frac{t}{T} \bigg) \bigg]$

We get, $ \lambda = \large\frac{2 \pi}{ 0.5 \times 10^3\: m }$$= 1.26 \: cm$

and $ \large\frac{1}{T}$ $ = Frequency =\large\frac{1.5 \times 10^{11} }{2 \pi}$$ = 23.9 \: GHz$

Ans : (b)

Ask Question

Tag:MathPhyChemBioOther

Take Test

...