Question 102: EM waves

Question

The magnetic field vector of an electromagnetic wave is given by $$B = {B_0}{{\widehat i + \widehat j} \over {\sqrt 2 }}\cos (kz - \omega t)$$; where $$\widehat i,\widehat j$$ represents unit vector along x and y-axis respectively. At t = 0s, two electric charges q1 of 4$$\pi$$ coulomb and q2 of 2$$\pi$$ coulomb located at $$\left( {0,0,{\pi  \over k}} \right)$$ and $$\left( {0,0,{{3\pi } \over k}} \right)$$, respectively, have the same velocity of 0.5 c $$\widehat i$$, (where c is the velocity of light). The ratio of the force acting on charge q1 to q2 is :-

Accepted Answer

$$\overrightarrow F  = q\left( {\overrightarrow V  	imes \overrightarrow B } ight)$$
$${\overrightarrow F _1} = 4\pi \left[ {0.5c\widehat i 	imes {B_0}\left( {{{\widehat i + \widehat j} \over 2}} ight)\cos \left( {K.{\pi  \over K} - 0} ight)} ight]$$
$${\overrightarrow F _2} = 2\pi \left[ {0.5c\widehat i 	imes {B_0}\left( {{{\widehat i + \widehat j} \over 2}} ight)\cos \left( {K.{{3\pi } \over K} - 0} ight)} ight]$$
$$\cos \pi  =  - 1$$, $$\cos 3\pi  =  - 1$$
$$	herefore$$ $${{{F_1}} \over {{F_2}}} = 2$$

Answer

$$2\sqrt 2 :1$$

Answer

$$1:\sqrt 2 $$

Answer

2 : 1

Answer

$$\sqrt 2 :1$$

EM waves

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