JEE MAIN - Physics (2019 - 11th January Morning Slot - No. 23)
In an experiment, electrons are accelerated, from rest, by applying a voltage of 500 V. Calculate the radius of the path if a magnetic field 100 mT is then applied. [Charge of the electron = 1.6 $$ \times $$ 10–19 C Mass of the electron = 9.1 $$ \times $$ 10–31 kg]
7.5 $$ \times $$ 10$$-$$4 m
7.5 $$ \times $$ 10$$-$$3 m
7.5 m
7.5 $$ \times $$ 10$$-$$2 m
Explanation
$$r = {{\sqrt {2mk} } \over {eB}} = {{\sqrt {2me\Delta v} } \over {eB}}$$
$$r = {{\sqrt {{{2m} \over e}.\Delta v} } \over B} = {{\sqrt {{{2 \times 9.1 \times {{10}^{ - 31}}} \over {1.6 \times {{10}^{ - 19}}}}\left( {500} \right)} } \over {100 \times {{10}^{ - 3}}}}$$
$$r = {{\sqrt {{{9.1} \over {0.16}} \times {{10}^{ - 10}}} } \over {{{10}^{ - 1}}}} = {3 \over 4} \times {10^{ - 4}}$$
$$ = 7.5 \times {10^{ - 4}}$$
$$r = {{\sqrt {{{2m} \over e}.\Delta v} } \over B} = {{\sqrt {{{2 \times 9.1 \times {{10}^{ - 31}}} \over {1.6 \times {{10}^{ - 19}}}}\left( {500} \right)} } \over {100 \times {{10}^{ - 3}}}}$$
$$r = {{\sqrt {{{9.1} \over {0.16}} \times {{10}^{ - 10}}} } \over {{{10}^{ - 1}}}} = {3 \over 4} \times {10^{ - 4}}$$
$$ = 7.5 \times {10^{ - 4}}$$
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