JEE MAIN - Physics (2019 - 11th January Evening Slot - No. 27)
The region between y = 0 and y = d contains a magnetic field $$\overrightarrow B = B\widehat z$$. A particle of mass m and charge q enters the region with a velocity $$\overrightarrow v = v\widehat i.$$ If d $$=$$ $${{mv} \over {2qB}},$$ the acceleration of the charged particle at the point of its emergence at the other side is :
$${{qvB} \over m}\left( -{{{\sqrt 3 } \over 2}\widehat i - {1 \over 2}\widehat j} \right)$$
$${{qvB} \over m}\left( {{1 \over 2}\widehat i - {{\sqrt 3 } \over 2}\widehat j} \right)$$
$${{qvB} \over m}\left( {{{ - \widehat j + \widehat i} \over {\sqrt 2 }}} \right)$$
$${{qvB} \over m}\left( {{{\widehat j + \widehat i} \over {\sqrt 2 }}} \right)$$
Explanation
_11th_January_Evening_Slot_en_27_1.png)
Here R = $${{mv} \over {qB}}$$ = 2d
cos $$\theta $$ = $${{{R \over 2}} \over R}$$ = $${1 \over 2}$$
$$ \Rightarrow $$ $$\theta $$ = 60o
Acceleration of the charged particle at the point of its emergence,
$$\overrightarrow {{a_c}} = {a_{{c_x}}}\left( { - \widehat i} \right) + {a_{{c_y}}}\left( { - \widehat j} \right)$$
= $${a_c}\cos 30^\circ \left( { - \widehat i} \right) + {a_c}\sin 30^\circ \left( { - \widehat j} \right)$$
= $${a_c}\left( {{{\sqrt 3 } \over 2}\left( { - \widehat i} \right) + {1 \over 2}\left( { - \widehat j} \right)} \right)$$
= $${{qvB} \over m}\left( { - {{\sqrt 3 } \over 2}\widehat i - {1 \over 2}\widehat j} \right)$$
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