JEE MAIN - Physics (2021 - 31st August Evening Shift - No. 12)
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 :-
$$2\sqrt 2 :1$$
$$1:\sqrt 2 $$
2 : 1
$$\sqrt 2 :1$$
Explanation
$$\overrightarrow F = q\left( {\overrightarrow V \times \overrightarrow B } \right)$$
$${\overrightarrow F _1} = 4\pi \left[ {0.5c\widehat i \times {B_0}\left( {{{\widehat i + \widehat j} \over 2}} \right)\cos \left( {K.{\pi \over K} - 0} \right)} \right]$$
$${\overrightarrow F _2} = 2\pi \left[ {0.5c\widehat i \times {B_0}\left( {{{\widehat i + \widehat j} \over 2}} \right)\cos \left( {K.{{3\pi } \over K} - 0} \right)} \right]$$
$$\cos \pi = - 1$$, $$\cos 3\pi = - 1$$
$$\therefore$$ $${{{F_1}} \over {{F_2}}} = 2$$
$${\overrightarrow F _1} = 4\pi \left[ {0.5c\widehat i \times {B_0}\left( {{{\widehat i + \widehat j} \over 2}} \right)\cos \left( {K.{\pi \over K} - 0} \right)} \right]$$
$${\overrightarrow F _2} = 2\pi \left[ {0.5c\widehat i \times {B_0}\left( {{{\widehat i + \widehat j} \over 2}} \right)\cos \left( {K.{{3\pi } \over K} - 0} \right)} \right]$$
$$\cos \pi = - 1$$, $$\cos 3\pi = - 1$$
$$\therefore$$ $${{{F_1}} \over {{F_2}}} = 2$$
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