JEE MAIN - Physics (2020 - 9th January Morning Slot - No. 10)
A charged particle of mass 'm' and charge 'q'
moving under the influence of uniform electric
field $$E\overrightarrow i $$
and a uniform magnetic field $$B\overrightarrow k $$
follows a trajectory from point P to Q as shown
in figure. The velocities at P and Q are
respectively, $$v\overrightarrow i $$ and $$ - 2v\overrightarrow j $$
. Then which of the
following statements (A, B, C, D) are the
correct ?
(Trajectory shown is schematic and not to scale) :_9th_January_Morning_Slot_en_10_1.png)
(A) E = $${3 \over 4}\left( {{{m{v^2}} \over {qa}}} \right)$$
(B) Rate of work done by the electric field at P is $${3 \over 4}\left( {{{m{v^3}} \over a}} \right)$$
(C) Rate of work done by both the fields at Q is zero
(D) The difference between the magnitude of angular momentum of the particle at P and Q is 2mav.
(Trajectory shown is schematic and not to scale) :
(A) E = $${3 \over 4}\left( {{{m{v^2}} \over {qa}}} \right)$$
(B) Rate of work done by the electric field at P is $${3 \over 4}\left( {{{m{v^3}} \over a}} \right)$$
(C) Rate of work done by both the fields at Q is zero
(D) The difference between the magnitude of angular momentum of the particle at P and Q is 2mav.
(A), (B), (C), (D)
(A), (B), (C)
(A), (C), (D)
(B), (C), (D)
Explanation
For the particle moving from point P to point Q,
WElectric field + WMagnetic field = $$\Delta $$KE
$$ \Rightarrow $$ qE(2$$a$$) + 0 = $${1 \over 2}m\left[ {{{\left( {2v} \right)}^2} - {v^2}} \right]$$
$$ \Rightarrow $$ 2qE$$a$$ = $${3 \over 2}m{v^2}$$
$$ \Rightarrow $$ E = $${3 \over 4}\left( {{{m{v^2}} \over {qa}}} \right)$$
$$ \therefore $$ Point (A) is correct.
Rate of work done(Power) by electric
field at p = $$\overrightarrow F .\overrightarrow v $$ = qEv = $${3 \over 4}\left( {{{m{v^3}} \over a}} \right)$$
$$ \therefore $$ Point (B) is also correct.
Angle between electric force and velocity is 90º, hence rate of work done will be zero at Q.
$$ \therefore $$ Point (C) is also correct.
Initial angular momentum at P, $${\overrightarrow L }$$i = mv$$a\left( { - \widehat k} \right)$$
Final angular momentum at Q, $${\overrightarrow L }$$f = m(2v)$$(2a)\left( { - \widehat k} \right)$$
Change in angular momentum $${\overrightarrow L }$$f – $${\overrightarrow L }$$i = $$3mva\left( { - \widehat k} \right)$$
Magnitude in the change in angular momentum $$\left| {\Delta \overrightarrow L } \right|$$ = $$3mva$$
$$ \therefore $$ Point (D) is wrong.
WElectric field + WMagnetic field = $$\Delta $$KE
$$ \Rightarrow $$ qE(2$$a$$) + 0 = $${1 \over 2}m\left[ {{{\left( {2v} \right)}^2} - {v^2}} \right]$$
$$ \Rightarrow $$ 2qE$$a$$ = $${3 \over 2}m{v^2}$$
$$ \Rightarrow $$ E = $${3 \over 4}\left( {{{m{v^2}} \over {qa}}} \right)$$
$$ \therefore $$ Point (A) is correct.
Rate of work done(Power) by electric
field at p = $$\overrightarrow F .\overrightarrow v $$ = qEv = $${3 \over 4}\left( {{{m{v^3}} \over a}} \right)$$
$$ \therefore $$ Point (B) is also correct.
Angle between electric force and velocity is 90º, hence rate of work done will be zero at Q.
$$ \therefore $$ Point (C) is also correct.
Initial angular momentum at P, $${\overrightarrow L }$$i = mv$$a\left( { - \widehat k} \right)$$
Final angular momentum at Q, $${\overrightarrow L }$$f = m(2v)$$(2a)\left( { - \widehat k} \right)$$
Change in angular momentum $${\overrightarrow L }$$f – $${\overrightarrow L }$$i = $$3mva\left( { - \widehat k} \right)$$
Magnitude in the change in angular momentum $$\left| {\Delta \overrightarrow L } \right|$$ = $$3mva$$
$$ \therefore $$ Point (D) is wrong.
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