JEE MAIN - Physics (2020 - 2nd September Evening Slot - No. 11)
The figure shows a region of length ‘l’ with a
uniform magnetic field of 0.3 T in it and a
proton entering the region with velocity 4 $$ \times $$ 105
ms–1 making an angle 60o with the field. If the
proton completes 10 revolution by the time it
cross the region shown, ‘l’ is close to
(mass of proton = 1.67 $$ \times $$ 10–27 kg, charge of the proton = 1.6 $$ \times $$ 10–19 C)_2nd_September_Evening_Slot_en_11_1.png)
(mass of proton = 1.67 $$ \times $$ 10–27 kg, charge of the proton = 1.6 $$ \times $$ 10–19 C)
0.22 m
0.11 m
0.88 m
0.44 m
Explanation
$$l$$ = 10 × pitch
= 10 $$ \times $$ vcos60o $$ \times $$ $${{2\pi m} \over {qB}}$$
= 10 $$ \times $$ v $$ \times $$ $${1 \over 2}$$ $$ \times $$ $${{2\pi m} \over {qB}}$$
= $${{10 \times 4 \times {{10}^5} \times 3.14 \times 1.67 \times {{10}^{ - 27}}} \over {1.6 \times {{10}^{ - 19}} \times 0.3}}$$
= 0.44 m
= 10 $$ \times $$ vcos60o $$ \times $$ $${{2\pi m} \over {qB}}$$
= 10 $$ \times $$ v $$ \times $$ $${1 \over 2}$$ $$ \times $$ $${{2\pi m} \over {qB}}$$
= $${{10 \times 4 \times {{10}^5} \times 3.14 \times 1.67 \times {{10}^{ - 27}}} \over {1.6 \times {{10}^{ - 19}} \times 0.3}}$$
= 0.44 m
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