JEE MAIN - Physics (2020 - 6th September Morning Slot - No. 11)
An electron is moving along +x direction with a velocity of 6 $$ \times $$ 106
ms–1. It enters a region of uniform
electric field of 300 V/cm pointing along +y direction. The magnitude and direction of the magnetic
field set up in this region such that the electron keeps moving along the x direction will be :
3 $$ \times $$ 10–4 T, along –z direction
5 $$ \times $$ 10–3 T, along –z direction
5 $$ \times $$ 10–3 T, along +z direction
3 $$ \times $$ 10–4 T, along +z direction
Explanation
_6th_September_Morning_Slot_en_11_1.png)
$$\overrightarrow B $$ must be in +z axis.
$$\overrightarrow V = 6 \times {10^6}\widehat i$$
$$\overrightarrow E = 300\widehat j$$ V/cm = 3 $$ \times $$ 104 $$\widehat j$$ V/m
$$\overrightarrow F $$ = $$q\overrightarrow E + q\overrightarrow V \times \overrightarrow B $$
$$ \therefore $$ $$q\overrightarrow E + q\overrightarrow V \times \overrightarrow B $$ = 0
$$ \Rightarrow $$qE = qVB
$$ \Rightarrow $$ B = $${E \over V}$$ = $${{3 \times {{10}^4}} \over {6 \times {{10}^6}}}$$ = 5 $$ \times $$ 10–3 T
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