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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)
View written solutionFree

Correct answer: D

To move in a straight line with constant velocity, the net Lorentz force must be zero:

F⃗=q(E⃗+v⃗×B⃗)=0\vec F = q(\vec E + \vec v \times \vec B)=0F=q(E+v×B)=0

So we must check each option and see whether

E⃗+v⃗×B⃗=0\vec E + \vec v \times \vec B = 0E+v×B=0

for the given particle. Note that if this vector is zero, then it is zero irrespective of the sign of charge, so electron/proton matters only if we were writing force explicitly.

1. Check Option A: (III)(ii)(R)(III)(ii)(R)(III)(ii)(R)

Given:

  • Particle: Proton with 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^​

Since v⃗=0\vec v=0v=0,

v⃗×B⃗=0\vec v \times \vec B = 0v×B=0

Hence,

E⃗+v⃗×B⃗=−E0y^≠0\vec E + \vec v \times \vec B = -E_0\hat y \neq 0E+v×B=−E0​y^​=0

So net force is not zero.

❌ Option A is incorrect.

2. Check Option B: (IV)(i)(S)(IV)(i)(S)(IV)(i)(S)

Given:

  • Particle: Proton with v⃗=2E0B0x^\vec v = 2\frac{E_0}{B_0}\hat xv=2B0​E0​​x^
  • E⃗=E0z^\vec E=E_0\hat zE=E0​z^
  • B⃗=B0z^\vec B=B_0\hat zB=B0​z^

Now compute:

v⃗×B⃗=2E0B0x^×B0z^=2E0(x^×z^)\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)v×B=2B0​E0​​x^×B0​z^=2E0​(x^×z^)

Using

x^×z^=−y^\hat x \times \hat z = -\hat yx^×z^=−y^​

we get

v⃗×B⃗=−2E0y^\vec v \times \vec B = -2E_0\hat yv×B=−2E0​y^​

Therefore,

E⃗+v⃗×B⃗=E0z^−2E0y^≠0\vec E + \vec v \times \vec B = E_0\hat z - 2E_0\hat y \neq 0E+v×B=E0​z^−2E0​y^​=0

So force is not zero.

❌ Option B is incorrect.

3. Check Option C: (III)(iii)(P)(III)(iii)(P)(III)(iii)(P)

Given:

  • Particle: Proton with v⃗=0\vec v=0v=0
  • E⃗=−E0x^\vec E=-E_0\hat xE=−E0​x^
  • B⃗=−B0x^\vec B=-B_0\hat xB=−B0​x^

Again, since v⃗=0\vec v=0v=0,

v⃗×B⃗=0\vec v \times \vec B = 0v×B=0

Thus,

E⃗+v⃗×B⃗=−E0x^≠0\vec E + \vec v \times \vec B = -E_0\hat x \neq 0E+v×B=−E0​x^=0

So the particle will accelerate due to electric field.

❌ Option C is incorrect.

4. Check Option D: (II)(iii)(S)(II)(iii)(S)(II)(iii)(S)

Given:

  • Particle: Electron with 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=B0z^\vec B=B_0\hat zB=B0​z^

Now compute magnetic term:

v⃗×B⃗=E0B0y^×B0z^=E0(y^×z^)\vec v \times \vec B = \frac{E_0}{B_0}\hat y \times B_0\hat z = E_0(\hat y \times \hat z)v×B=B0​E0​​y^​×B0​z^=E0​(y^​×z^)

Using

y^×z^=x^\hat y \times \hat z = \hat xy^​×z^=x^

we get

v⃗×B⃗=E0x^\vec v \times \vec B = E_0\hat xv×B=E0​x^

Therefore,

E⃗+v⃗×B⃗=−E0x^+E0x^=0\vec E + \vec v \times \vec B = -E_0\hat x + E_0\hat x = 0E+v×B=−E0​x^+E0​x^=0

Hence,

F⃗=q(0)=0\vec F = q(0)=0F=q(0)=0

So the particle continues in a straight line with constant velocity.

✅ Option D is correct.

Final Answer

The particle moves in a straight line with constant velocity only in:

(II)(iii)(S)(II)(iii)(S)(II)(iii)(S)

So the correct option is D.

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