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A positive charge particle of mass m and...

A positive charge particle of mass m and charge q is projected with velocity v as shown in Fig. If radius of curvature of charge particle in magnetic field region is greater than d, then find the time spent by the charge particle in magnetic field.

A

`(2m)/(qB) sin^(-1)((dqB)/(mv))`

B

`(2m)/(qB) sin^(-1)((mv)/(dqB))`

C

`(m)/(qB) sin^(-1)((dqB)/(mv))`

D

`(m)/(qB) sin^(-1)((mv)/(dqB))`

Text Solution

Verified by Experts

The correct Answer is:
C
`R=(mv)/(qB), sin theta=d/R=(dqB)/(mv)`
`t=(2 pi m)/(qB) (theta)/(qB) theta=(m)/(qB) sin^(_1) [(dqB)/(mv)]`
.
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Knowledge Check

  • There exist uniform magnetic field vec(B)=-B_(0)hatk in region of space with 0ltxltd and 2dltxlt3d as shown in the figure. A positive charged particle of mass m and charge q is projected with velocity vec(v)=vhat(i) as shown in the figure. If radius of curvature of path of the charged particle in magnetic field is R(2dltRlt3d) then time elapse by charged particle in magnetic field regions is

    A
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    B
    `m/(qB)sin^(-1)((2d)/R)`
    C
    `m/(qB)`
    D
    `m/(qB)sin^(-1)(d/R)`

  • There exist uniform magnetic field vec(B)=-B_(0)hatk in region of space with 0ltxltd and 2dltxlt3d as shown in the figure. A positive charged particle of mass m and charge q is projected with velocity vec(v)=vhat(i) as shown in the figure. If radius of curvature of path of the charged particle in magnetic field is R(2dltRlt3d) then time elapse by charged particle in magnetic field regions is

    A
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    C
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    D
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  • The radius of curvature of the path of a charged particle in a uniform magnetic field is directly proportional to

    A
    the charge on the particle
    B
    the momentum of the particle
    C
    the energy of the particle
    D
    the intensity of the field

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