AB and AC are the boundary lines within which a magnetic field B exists.If the magnetic field is absent, a charged particle of mass m and charge q must have passed through a point P on angle bisector of —BAC, at a distance r 2from A, if it has fallen on AB normally at a point Q such that OP = r. If the magnetic field is present, how much time the charged particle will take to come out of the magnetic field

Since, QP=r,AP=r2 and ∠AQP=90° AQ=r and ∠QAP=45° The same logic concludes AR=r and ∠PAR=45° So, ∠QAR=90° r=mvBq Distance travelled =πr2 Time =π2·mvBq·1v=mπ2Bq

AB and AC are the boundary lines within which a magnetic field B exists.If the magnetic field is absent, a charged particle of mass m and charge q must have passed through a point P on angle bisector of —BAC, at a distance r 2from A, if it has fallen on AB normally at a point Q such that OP = r. If the magnetic field is present, how much time the charged particle will take to come out of the magnetic field

Since, QP=r,AP=r2 and ∠AQP=90° AQ=r and ∠QAP=45° The same logic concludes AR=r and ∠PAR=45° So, ∠QAR=90° r=mvBq Distance travelled =πr2 Time =π2·mvBq·1v=mπ2Bq

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AB and AC are the boundary lines within which a magnetic field B exists.
If the magnetic field is absent, a charged particle of mass m and charge q must have passed through a point P on angle bisector of —BAC, at a distance r $\sqrt{2}$ from A, if it has fallen on AB normally at a point Q such that OP = r. If the magnetic field is present, how much time the charged particle will take to come out of the magnetic field

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$\frac{m π}{B q}$

$\frac{4 m π}{B q}$

$\frac{2 m π}{B q}$

$\frac{m π}{2 B q}$

answer is D.

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Detailed Solution

$Since, Q P = r, A P = r \sqrt{2} and ∠ A Q P = 90 ° A Q = r and ∠ Q A P = 45 ° The same logic concludes A R = r and ∠ P A R = 45 ° So, ∠ Q A R = 90 ° r = \frac{m v}{B q} Distance travelled = \frac{π r}{2} Time = \frac{π}{2} \cdot \frac{m v}{B q} \cdot \frac{1}{v} = \frac{m π}{2 B q}$