The magnetic field of a plane electromagnetic wave is given by :
$$\vec B = {B_0}\hat i\left[ {\cos \left( {kz - \omega t} \right)} \right] + {B_1}\hat j\cos \left( {kz + \omega t} \right)$$
Where $${B_0} = 3 \times {10^{ - 5}}T$$    and $${B_1} = 2 \times {10^{ - 6}}T.$$
The $$rms$$  value of the force experienced by a stationary charge $$Q = {10^{ - 4}}C$$   at $$z = 0$$  is closest to:

Solution :
$$\eqalign{ & {B_0} = \sqrt {B_0^2 + B_1^2} = \sqrt {{{30}^2} + {2^2}} \times {10^{ - 6}} \cr & \approx 30 \times {10^{ - 6}}T \cr & \therefore {E_0} = cB = 3 \times {10^8} \times 30 \times {10^{ - 6}} \cr & = 9 \times {10^3}V/m \cr & {E_{rms}} = \frac{{{E_0}}}{{\sqrt 2 }} = \frac{9}{{\sqrt 2 }} \times {10^3}V/m \cr} $$
Force on the charge,
$$F = {E_{rms}}Q = \frac{9}{{\sqrt 2 }} \times {10^3} \times {10^{ - 4}} \simeq 0.64N$$