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Magnetism question

2017 · Shift 1 · Q50
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Magnetism question

2017 · Shift 1 · Q50

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 would the particle move in a straight line along the negative direction of yyy-axis (i.e., move along −y^- \widehat y−y​)?
  1. A
    (II)(iii)(Q)\left( {{\rm I}{\rm I}} \right)\left( {iii} \right)\left( Q \right)(II)(iii)(Q)
  2. B
    (III)(ii)(R)\left( {{\rm I}{\rm I}{\rm I}} \right)\left( {ii} \right)\left( R \right)(III)(ii)(R)
  3. C
    (IV)(ii)(S)\left( {{\rm I}V} \right)\left( {ii} \right)\left( S \right)(IV)(ii)(S)
  4. D
    (III)(ii)(P)\left( {{\rm I}{\rm I}{\rm I}} \right)\left( {ii} \right)\left( P \right)(III)(ii)(P)
View written solutionFree

Correct answer: B

  1. Condition for motion in a straight line along \(-\hat y\)

For a charged particle to move in a straight line, its acceleration must remain along the direction of motion and there should be no transverse deflection.

The Lorentz force is

F⃗=q(E⃗+v⃗×B⃗).\vec F = q(\vec E + \vec v \times \vec B).F=q(E+v×B).

We need the particle to move along \(-\hat y\).

So we check each option and see whether the net force is along \(-\hat y\)** and whether the initial velocity is compatible.

  1. Option A: \((II)(iii)(Q)\)

Given:

  • Electron
  • v⃗=E0B0y^\vec v = \frac{E_0}{B_0}\hat yv=B0​E0​​y^​
  • E⃗=−E0x^\vec E = -E_0\hat xE=−E0​x^
  • B⃗=B0x^\vec B = B_0\hat xB=B0​x^

First, velocity is initially along \(+\hat y\), not \(-\hat y\). So already suspicious.

Now,

v⃗×B⃗=E0B0y^×B0x^=E0(y^×x^)=−E0z^.\vec v \times \vec B = \frac{E_0}{B_0}\hat y \times B_0\hat x = E_0(\hat y \times \hat x) = -E_0\hat z.v×B=B0​E0​​y^​×B0​x^=E0​(y^​×x^)=−E0​z^.

Thus,

E⃗+v⃗×B⃗=−E0x^−E0z^.\vec E + \vec v \times \vec B = -E_0\hat x - E_0\hat z.E+v×B=−E0​x^−E0​z^.

For electron, \(q=-e\), so force is along

F⃗=−e(−E0x^−E0z^)=eE0x^+eE0z^,\vec F = -e(-E_0\hat x - E_0\hat z)= eE_0\hat x + eE_0\hat z,F=−e(−E0​x^−E0​z^)=eE0​x^+eE0​z^,

which is not along \(-\hat y\).

So A is incorrect.

  1. Option B: \((III)(ii)(R)\)

Given:

  • Proton
  • v⃗=0\vec v = 0v=0
  • E⃗=−E0y^\vec E = -E_0\hat yE=−E0​y^​
  • B⃗=B0y^\vec B = B_0\hat yB=B0​y^​

Initially,

v⃗×B⃗=0.\vec v \times \vec B = 0.v×B=0.

So force is only electric:

F⃗=qE⃗=e(−E0y^)=−eE0y^.\vec F = q\vec E = e(-E_0\hat y) = -eE_0\hat y.F=qE=e(−E0​y^​)=−eE0​y^​.

Thus proton starts accelerating along \(-\hat y\).

As it moves, its velocity remains along \(-\hat y\), while magnetic field is along \(+\hat y\). Since they are antiparallel,

v⃗×B⃗=0\vec v \times \vec B = 0v×B=0

at all times.

Hence the particle continues to move in a straight line along \(-\hat y\).

So B is correct.

  1. Option C: \((IV)(ii)(S)\)

Given:

  • Proton
  • v⃗=2E0B0x^\vec v = 2\frac{E_0}{B_0}\hat xv=2B0​E0​​x^
  • E⃗=−E0y^\vec E = -E_0\hat yE=−E0​y^​
  • B⃗=B0z^\vec B = B_0\hat zB=B0​z^

Compute magnetic term:

v⃗×B⃗=2E0B0x^×B0z^=2E0(x^×z^)=−2E0y^.\vec v \times \vec B = 2\frac{E_0}{B_0}\hat x \times B_0\hat z = 2E_0(\hat x \times \hat z) = -2E_0\hat y.v×B=2B0​E0​​x^×B0​z^=2E0​(x^×z^)=−2E0​y^​.

Hence,

E⃗+v⃗×B⃗=−E0y^−2E0y^=−3E0y^.\vec E + \vec v \times \vec B = -E_0\hat y -2E_0\hat y = -3E_0\hat y.E+v×B=−E0​y^​−2E0​y^​=−3E0​y^​.

So force is along \(-\hat y\), but the initial velocity is along \(+\hat x\). Therefore the particle will not move in a straight line along the negative y-axis from the start; it will have x-motion too.

So C is incorrect.

  1. Option D: \((III)(ii)(P)\)

Given:

  • Proton
  • v⃗=0\vec v = 0v=0
  • E⃗=−E0y^\vec E = -E_0\hat yE=−E0​y^​
  • B⃗=−B0x^\vec B = -B_0\hat xB=−B0​x^

Initially, since \(\vec v=0\), magnetic force is zero, and electric force is

F⃗=e(−E0y^)=−eE0y^.\vec F = e(-E_0\hat y) = -eE_0\hat y.F=e(−E0​y^​)=−eE0​y^​.

So it starts moving along \(-\hat y\).

But once velocity becomes along \(-\hat y\), magnetic force appears:

v⃗×B⃗∝(−y^)×(−x^)=y^×x^=−z^.\vec v \times \vec B \propto (-\hat y) \times (-\hat x)= \hat y \times \hat x = -\hat z.v×B∝(−y^​)×(−x^)=y^​×x^=−z^.

So magnetic force is nonzero and perpendicular to motion. Hence the particle will bend and not continue in a straight line.

So D is incorrect.

  1. Final conclusion

Only Option B satisfies the condition that the particle moves in a straight line along \(-\hat y\).

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