Home
Class 12
PHYSICS
A proton takes 10^(-12)s to complete one...

A proton takes `10^(-12)s` to complete one revolution in uniform magnetic field. The time taken in another orbit of double the radius in the same field is

A

`0.5 xx 10^(-12) sec `

B

`2 xx 10^(-12) sec `

C

`4 xx 10^(-12) sec `

D

`10^(-12) sec `

Text Solution

Verified by Experts

The correct Answer is:
D
`T_1/T_2 = m_1/m_2 xx q_2/q_1`
Promotional Banner

Topper's Solved these Questions

  • MOVING CHARGES AND MAGNETISM

    NARAYNA|Exercise EXERCISE - 1(H.W) (CYCLOTRON)|2 Videos

  • MOVING CHARGES AND MAGNETISM

    NARAYNA|Exercise EXERCISE - 1(H.W) (FORCE ON CURRENT CARRYING CONDUCTOR IN A MAGNETIC FIELD)|1 Videos

  • MOVING CHARGES AND MAGNETISM

    NARAYNA|Exercise EXERCISE - 1(H.W) (SOLENOID AND TOROID)|1 Videos

  • MAGNETISM AND MATTER

    NARAYNA|Exercise EXERCISE - 4 (SINGLE ANSWER TYPE QUESTION)|17 Videos

  • NUCLEAR PHYSICS

    NARAYNA|Exercise LEVEL-II-(H.W)|9 Videos

Similar Questions

Explore conceptually related problems

A protons of mass m charge 2 e is released from rest in a uniform electric field of strength E .The time taken by it to travel a distance d in the field is..

A protons of mass m charge c is released form rest in a uniform electric field of strength E .The time taken by it to travel a distance d in the field is

Time taken by an electrons to complete one revolution in the Bohr orbit of the H atom is

A proton moving with a velocity of 2xx10^(6)ms^(-1) describes circle of radius R in a magnetic field. The speed of an alpha- particle to describe a circle of same radius in the same magnitude field is

Protons are shot in a region perpendicular to a uniform magnetic field in a region :

A beam of protons with a velocity 4 xx 10^(5) m/sec enters a uniform magnetic field of 0.3 Testa at an angle of 60^(@) to the magnetic field. Find the radius of the helical path taken by the proton beam. Also find the pitch of the helix, which is the distance travelled by a proton in the beam parallel to the magnetic field during one period of rotation [Mass of proton = 1.67 xx 10^(-27) kg. charge on proton = 1.6 xx 10^(19)C ]

A proton of mass m and charge q enters a region of uniform magnetic field of a magnitude B with a velocity v directed perpendicular to the magnetic field. It moves in a circular path and leaves the magnetic field after completing a quarter of a circle. The time spent by the proton inside the magnetic field is proportional to: