JEE MAIN - Physics (2019 - 12th April Evening Slot - No. 4)
An electron, moving along the x-axis with an initial energy of 100 eV, enters a region of magnetic field $$\overrightarrow B = \left( {1.5 \times {{10}^{ - 3}}T} \right)\widehat k$$
at S (See figure). The field extends between x = 0 and x = 2 cm. The electron is
detected at the point Q on a screen placed 8 cm away from the point S. The distance d between P and Q (on
the screen) is :
(electron’s charge = 1.6 × 10–19 C, mass of electron = 9.1 × 10–31 kg)
_12th_April_Evening_Slot_en_4_2.png)
2.25 cm
12.87 cm
1.22 cm
11.65 cm
Explanation
We know, R = $${{mv} \over {qB}}$$
= $${{\sqrt {2mK} } \over {qB}}$$
= $${{\sqrt {2 \times 9.1 \times {{10}^{ - 31}} \times 100 \times 1.6 \times {{10}^{ - 19}}} } \over {1.6 \times {{10}^{ - 19}} \times 1.5 \times {{10}^{ - 3}}}}$$
= $${\sqrt 5 }$$ cm_12th_April_Evening_Slot_en_4_4.png)
From diagram,
$$\sin \theta = {2 \over R}$$ = $${2 \over {\sqrt 5 }}$$
So $$\tan \theta = {2 \over 1}$$
In $$\Delta $$MNQ,
$$\tan \theta = {x \over 6}$$
$$ \therefore $$ $${x \over 6}$$ = $${2 \over 1}$$
$$ \Rightarrow $$ $$x$$ = 12 cm
From diagram you can see,
QP = d > 12
By checking all the options you can see possible value of d = 12.87 cm
= $${{\sqrt {2mK} } \over {qB}}$$
= $${{\sqrt {2 \times 9.1 \times {{10}^{ - 31}} \times 100 \times 1.6 \times {{10}^{ - 19}}} } \over {1.6 \times {{10}^{ - 19}} \times 1.5 \times {{10}^{ - 3}}}}$$
= $${\sqrt 5 }$$ cm
From diagram,
$$\sin \theta = {2 \over R}$$ = $${2 \over {\sqrt 5 }}$$
So $$\tan \theta = {2 \over 1}$$
In $$\Delta $$MNQ,
$$\tan \theta = {x \over 6}$$
$$ \therefore $$ $${x \over 6}$$ = $${2 \over 1}$$
$$ \Rightarrow $$ $$x$$ = 12 cm
From diagram you can see,
QP = d > 12
By checking all the options you can see possible value of d = 12.87 cm
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