JEE MAIN - Physics (2019 - 9th January Evening Slot - No. 21)
A particle having the same charge as of electron moves in a ciurcular path of radius 0.5 cm under the influence of a magnetic field 0f 0.5 T. If an electric field of 100 V/m makes it to move in a straight path, then the mass of the particle is (Given charge of electron = 1.6 $$ \times $$ 10$$-$$19C)
9.1 $$ \times $$ 10$$-$$31 kg
1.6 $$ \times $$ 10$$-$$27 kg
1.6 $$ \times $$ 10$$-$$19 kg
2.0 $$ \times $$ 10$$-$$24 kg
Explanation
Given,
radius of circular path(r) = 0.5 cm
Magnetic field (B) = 0.5 T
Electric field (E) = 100 V/m
Charge of particle (q) = 1.6$$ \times $$10$$-$$19 C
As particle is moving in a circular path so,
$${{m{v^2}} \over r} = qvB$$
$$ \Rightarrow $$ r = $${{mv} \over {qB}}$$ . . . . . . . (1)
When electric field of 100 v/m is applied on the particle then particle is moving in the straight line. So, the net force on the particle is zero.
$$ \therefore $$ Fnet = 0
$$ \Rightarrow $$ Fe = Fm
$$ \Rightarrow $$ qE = qvB
$$ \Rightarrow $$ E = vB . . . . .(2)
From equation (1) and (2) we get,
r = $${m \over {qB}}\left( {{E \over B}} \right)$$
= $${{mE} \over {q{B^2}}}$$
$$ \Rightarrow $$ m = $${{q{B^2}r} \over E}$$
= $${{1.6 \times {{10}^{ - 19}} \times {{\left( {0.5} \right)}^2} \times 0.5 \times {{10}^{ - 2}}} \over {100}}$$
= 2 $$ \times $$ 10$$-$$24 kg
radius of circular path(r) = 0.5 cm
Magnetic field (B) = 0.5 T
Electric field (E) = 100 V/m
Charge of particle (q) = 1.6$$ \times $$10$$-$$19 C
As particle is moving in a circular path so,
$${{m{v^2}} \over r} = qvB$$
$$ \Rightarrow $$ r = $${{mv} \over {qB}}$$ . . . . . . . (1)
When electric field of 100 v/m is applied on the particle then particle is moving in the straight line. So, the net force on the particle is zero.
$$ \therefore $$ Fnet = 0
$$ \Rightarrow $$ Fe = Fm
$$ \Rightarrow $$ qE = qvB
$$ \Rightarrow $$ E = vB . . . . .(2)
From equation (1) and (2) we get,
r = $${m \over {qB}}\left( {{E \over B}} \right)$$
= $${{mE} \over {q{B^2}}}$$
$$ \Rightarrow $$ m = $${{q{B^2}r} \over E}$$
= $${{1.6 \times {{10}^{ - 19}} \times {{\left( {0.5} \right)}^2} \times 0.5 \times {{10}^{ - 2}}} \over {100}}$$
= 2 $$ \times $$ 10$$-$$24 kg
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