Home
Class 12
PHYSICS
A charged particle with charge q enters ...

A charged particle with charge `q` enters a region of constant, uniform and mututally orthogonal fields `vec(E) and vec(B)` with a velocity `vec(v)` perpendicular to both `vec(E) and vec(B)`, and comes out without any change in magnitude or direction of `vec(v)`. Then

A

`vecv=vecB xx vecE//E^2`

B

`vecv=vecE xx vecB//B^2`

C

`vecv=vecB xx vecE//B^2`

D

`vecv=vecE xx vecB//E^2`

Text Solution

Verified by Experts

The correct Answer is:
B
When `vecE` and `vecB` are perpendicular and velocity has no change then `qE=qvB" i.e., "v=(E)/(B)`. The two forces oppose each other so, v is along `vecE xx vecB` i.e., `vecv=(vecExx vecB)/(B^(2))`
As `vecE and vecB` are perpendicular to each other
`(|vecE xx vecB|)/(B^(2))=(EB sin 90^(@))/(B^(2))=(E)/(B)`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • MOVING CHARGES AND MAGNETISM

    NCERT FINGERTIPS|Exercise Magnetic Field Due To A Current Element, Biot-Savart Law|4 Videos

  • MOVING CHARGES AND MAGNETISM

    NCERT FINGERTIPS|Exercise Magnetic Field On The Axis Of A Circular Current Loop|8 Videos

  • MOVING CHARGES AND MAGNETISM

    NCERT FINGERTIPS|Exercise Motion In Magnetic Field|6 Videos

  • MAGNETISM AND MATTER

    NCERT FINGERTIPS|Exercise Assertion And Reason|15 Videos

  • NUCLEI

    NCERT FINGERTIPS|Exercise Assertion And Reason|15 Videos

Similar Questions

Explore conceptually related problems

If vec c is perpendicular to both vec a and vec b, then prove that it is perpendicular to both vec a+vec b and vec a-vec b .

If vec c is perpendicular to both vec a and vec b, then prove that it is perpendicular to both vec a+vec b and vec a-vec b

Knowledge Check

  • A charged particle with charge q enters a region of constant, uniform and mutually orthogonal fields vec E and vec B with a velocity perpendicular to both vec E " and " vec B , and comes out without any change in magnitude or direction of vec v . Then

    A
    ` vec v = vec B xx vec E // E^2`
    B
    `vec v = vec E xx vec B // B^2`
    C
    `vec v = vec B xx vec E // B^2`
    D
    `vec v = vec E xx vec B // E^2`

  • If vec(v)_(1)+vec(v)_(2) is perpendicular to vec(v)_(1)-vec(v)_(2) , then

    A
    `vec(v)_(1)` is perpendicular to `vec(v)_(2)`.
    B
    `|vec(v)_(1)|=|vec(v)_(2)|`
    C
    `vec(v)_(1)` is a null vector
    D
    The angle between `vec(v)_(1)` and `vec(v)_(2)` can have any value

  • If a charged particle goes unaccelerated in a region containing electric and magnetic fields, (i) vec(E) must be perpendicular to vec(B) (ii) vec(v) must be perpendicular to vec(E) (iii) vec(v) must be perpendicular to vec(B) (iv) E must be equal to v B

    A
    (i), (ii)
    B
    (ii), (iv)
    C
    (ii), (iii)
    D
    (i), (iv)

    • Similar Questions

    Explore conceptually related problems

    If vec a+vec b is perpendicular to vec b and vec a+vec 2b is perpendicular to vec a then

    If vec(c ) is the unit vector perpendicular to both the vector vec(a) and vec(b) , then what is another unit vector perpendicular to both the vectors vec(a) and vec(b) ?

    Two vector vec(A) and vec(B) have equal magnitudes. Then the vector vec(A) + vec(B) is perpendicular :

    A charge particle moves with velocity vec(V) in a uniform magnetic field vec(B) . The magnetic force experienced by the particle is

    Force vec(F) acting on a test charge q_(0) in a uniform electric field vec(E) is

    Home
    Profile