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Magnetics question

2007 · Shift 0 · Q76
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  5. /2007 · Shift 0 · Q76

Magnetics question

2007 · Shift 0 · Q76

JEE MainPhysicsMagneticsMCQ+4 / −1
A charged particle with charge qqq enters a region of constant, uniform and mutually orthogonal fields E→\overrightarrow EE and B→\overrightarrow BB with a velocity v→\overrightarrow vv perpendicular to both E→\overrightarrow EE and B→,\overrightarrow B,B, and comes out without any change in magnitude or direction of v→\overrightarrow vv. Then
  1. A
    v→=B→×E→/E2\overrightarrow v = \overrightarrow B \times \overrightarrow E /{E^2}v=B×E/E2
  2. B
    v→=E→×B→/B2\overrightarrow v = \overrightarrow E \times \overrightarrow B /{B^2}v=E×B/B2
  3. C
    v→=B→×E→/B2\overrightarrow v = \overrightarrow B \times \overrightarrow E /{B^2}v=B×E/B2
  4. D
    v→=E→×B→/E2\overrightarrow v = \overrightarrow E \times \overrightarrow B /{E^2}v=E×B/E2
View written solutionFree

Correct answer: B

  1. Condition for no change in velocity

If the particle comes out with no change in magnitude or direction of v⃗\vec vv, then its acceleration must be zero inside the field region.

So, the net Lorentz force must be zero:

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

Since q≠0q\neq 0q=0,

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

or

v⃗×B⃗=−E⃗\vec v \times \vec B = -\vec Ev×B=−E
  1. Use the given mutual perpendicularity

Given:

  • E⃗⊥B⃗\vec E \perp \vec BE⊥B
  • v⃗⊥E⃗\vec v \perp \vec Ev⊥E
  • v⃗⊥B⃗\vec v \perp \vec Bv⊥B

So v⃗\vec vv must be along the direction of E⃗×B⃗\vec E \times \vec BE×B (or opposite to it).

Let us solve vectorially.

Take cross product of

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

with B⃗\vec BB:

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

Now use the identity:

(a⃗×b⃗)×c⃗=b⃗(a⃗⋅c⃗)−a⃗(b⃗⋅c⃗)(\vec a \times \vec b)\times \vec c = \vec b(\vec a\cdot \vec c)-\vec a(\vec b\cdot \vec c)(a×b)×c=b(a⋅c)−a(b⋅c)

So,

(v⃗×B⃗)×B⃗=B⃗(v⃗⋅B⃗)−v⃗(B2)(\vec v \times \vec B)\times \vec B = \vec B(\vec v\cdot \vec B)-\vec v(B^2)(v×B)×B=B(v⋅B)−v(B2)

Since v⃗⊥B⃗\vec v \perp \vec Bv⊥B, we have v⃗⋅B⃗=0\vec v\cdot \vec B=0v⋅B=0. Hence,

(v⃗×B⃗)×B⃗=−v⃗B2(\vec v \times \vec B)\times \vec B = -\vec v B^2(v×B)×B=−vB2

Therefore,

E⃗×B⃗−v⃗B2=0\vec E \times \vec B - \vec v B^2 = 0E×B−vB2=0

which gives

v⃗=E⃗×B⃗B2\vec v = \frac{\vec E \times \vec B}{B^2}v=B2E×B​
  1. Match with options

This is exactly Option B:

v⃗=E⃗×B⃗B2\boxed{\vec v = \frac{\vec E \times \vec B}{B^2}}v=B2E×B​​
  1. Check other options briefly

  • A: B⃗×E⃗E2=−E⃗×B⃗E2\dfrac{\vec B\times \vec E}{E^2}= -\dfrac{\vec E\times\vec B}{E^2}E2B×E​=−E2E×B​

    Wrong direction and wrong denominator.

  • B: E⃗×B⃗B2\dfrac{\vec E\times\vec B}{B^2}B2E×B​

    Correct.

  • C: B⃗×E⃗B2=−E⃗×B⃗B2\dfrac{\vec B\times\vec E}{B^2}= -\dfrac{\vec E\times\vec B}{B^2}B2B×E​=−B2E×B​

    Wrong direction.

  • D: E⃗×B⃗E2\dfrac{\vec E\times\vec B}{E^2}E2E×B​

    Wrong denominator.

  1. Final Answer
v⃗=E⃗×B⃗B2\boxed{\vec v = \frac{\vec E \times \vec B}{B^2}}v=B2E×B​​

So, the correct option is B.

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