આકૃતિમાં દર્શાવ્યા પ્રમાણે એક $x-$ અક્ષ પર $100\, eV$ ઉર્જાથી ગતિ કરતો ઈલેક્ટ્રોન $\vec B = (1.5\times10^{-3}T)\hat k$ જેટલા ચુંબકીય ક્ષેત્રમાં $S$ આગળથી દાખલ થાય છે.ચુંબકીય ક્ષેત્ર $x = 0$ અને $x = 2\, cm$ વચ્ચે પ્રવર્તે છે.$S$ બિંદુથી $8\, cm$ દૂરના પડદા પર $Q$ બિંદુ આગળ ઇલેક્ટ્રોન નોંધાય છે તો $P$ અને $Q$ વચ્ચેનું અંતર $d$ .....$cm$ હશે?

Question

A$1.22$
B$2.25$
C$12.87$
D$11.65$

JEE MAIN 2019, Diffcult

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Solution

c
${\text{R}} = \frac{{{\text{mv}}}}{{{\text{qB}}}}$

$ = \frac{{\sqrt {2{\text{m}}({\text{KE}})} }}{{{\text{qB}}}}$

${\text{R}} = \frac{{\sqrt {2 \times 9.1 \times {{10}^{ - 31}} \times \left( {100 \times 1.6 \times {{10}^{ - 18}}} \right)} }}{{1.6 \times {{10}^{ - 10}} \times 1.5 \times {{10}^{ - 3}}}}$

${\text{R}} = 2.248\,{\text{cm}}$

$\sin \theta = \frac{2}{{2.248}};\,\,\,\,\,\tan \,\theta = \frac{{QU}}{{TU}}$

$\frac{2}{{1.026}} = \frac{{QU}}{6}$

$QU = 11.69$

${\text{PU}} = {\text{R}}(1 - \cos \,\theta )$

$ = 1.22$

${\text{d}} = {\text{QU}} + {\text{PU}}$

$(i)$	$(ii)$	$(iii)$
(A) $\frac{{{\mu _0}i}}{r}$ $\otimes$	(A) $\frac{{{\mu _0}i}}{4}\left( {\frac{1}{{{r_1}}} - \frac{1}{{{r_2}}}} \right)$ $\otimes$	(A) $\frac{{{\mu _0}i}}{4}\left( {\frac{1}{{{r_1}}} - \frac{1}{{{r_2}}}} \right)$ $\otimes$
(B) $\frac{{{\mu _0}i}}{{2r}}$ $\odot$	(B) $\frac{{{\mu _0}i}}{4}\left( {\frac{1}{{{r_1}}} + \frac{1}{{{r_2}}}} \right)$ $\otimes$	(B) $\frac{{{\mu _0}i}}{4}\left( {\frac{1}{{{r_1}}} + \frac{1}{{{r_2}}}} \right)$ $\otimes$
(C) $\frac{{{\mu _0}i}}{{4r}}$ $\otimes$	(C) $\frac{{{\mu _0}i}}{4}\left( {\frac{1}{{{r_1}}} - \frac{1}{{{r_2}}}} \right)$ $\odot$	(C)$\frac{{{\mu _0}i}}{4}\left( {\frac{1}{{{r_1}}} - \frac{1}{{{r_2}}}} \right)$ $\odot$
(D) $\frac{{{\mu _0}i}}{{4r}}$ $\odot$	(D) $0$	(D) $0$

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