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Magnetism question

2017 · Shift 1 · Q37
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Magnetism question

2017 · Shift 1 · Q37

JEE AdvancedPhysicsMagnetismMCQ+3 / −0.75
A charged particle (electron or proton) is introduced at the origin (x=0,y=0,z=0) with a given initial velocity v→.\overrightarrow v .v. A uniform electric field E→\overrightarrow EE and a uniform magnetic field B→\overrightarrow BB exist everywhere. The velocity v→,\overrightarrow v ,v, electric field E→\overrightarrow EE and magnetic field B→\overrightarrow BB are given in column 1,21,21,2 and 3,3,3, respectively. The quantities E0,B0{E_0},{B_0}E0​,B0​ are positive in magnitude.

Column 1 Column 2 Column 3
(I) Electron with v→=2E0B0x^\overrightarrow v = 2{{{E_0}} \over {{B_0}}}\widehat xv=2B0​E0​​x (i) E→=E0z^\overrightarrow E = {E_0}\widehat zE=E0​z (P) B→=−B0x^\overrightarrow B = - {B_0}\widehat xB=−B0​x
(II) Electron with v→=E0B0y^\overrightarrow v = {{{E_0}} \over {{B_0}}}\widehat yv=B0​E0​​y​ (ii) E→=−E0y^\overrightarrow E = - {E_0}\widehat yE=−E0​y​ (Q) B→=B0x^\overrightarrow B = {B_0}\widehat xB=B0​x
(III) Proton with v→=0\overrightarrow v = 0v=0 (iii) E→=−E0x^\overrightarrow E = - {E_0}\widehat xE=−E0​x (R) B→=B0y^\overrightarrow B = {B_0}\widehat yB=B0​y​
(IV) Proton with v→=2E0B0x^\overrightarrow v = 2{{{E_0}} \over {{B_0}}}\widehat xv=2B0​E0​​x (iv) E→=E0x^\overrightarrow E = {E_0}\widehat xE=E0​x (S) B→=B0z^\overrightarrow B = {B_0}\widehat zB=B0​z
In which case will the particle move in a straight line with constant velocity?
  1. A
    (III)(ii)(R)\left( {{\rm I}{\rm I}{\rm I}} \right)\left( {ii} \right)\left( R \right)(III)(ii)(R)
  2. B
    (IV)(i)(S)\left( {{\rm I}V} \right)\left( i \right)\left( S \right)(IV)(i)(S)
  3. C
    (III)(iii)(P)\left( {{\rm I}{\rm I}{\rm I}} \right)\left( {iii} \right)\left( P \right)(III)(iii)(P)
  4. D
    (II)(iii)(S)\left( {{\rm I}{\rm I}} \right)\left( {iii} \right)\left( S \right)(II)(iii)(S)
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