Home
Class 12
PHYSICS
Ratio of electric and magnetic field du...

Ratio of electric and magnetic field due to moving point charge if its speed is `4.5xx10^(5)m//s`

A

`2xx10^(11)`

B

`3xx10^(11)`

C

`2xx10^(8)`

D

`3xx10^(12)`

Text Solution

Verified by Experts

The correct Answer is:
A
`E=(KQ)/(r^(2)), B=(mu_(o))/(4pi) (qV)/(r^(2))`
`(E )/(B)=(KQ.4pi)/(mu_(o)qV)=(1)/(4pi epsilon_(o))(4pi)/(mu_(o)) =(C^(2))/(V)=(9xx 10^(16))/(4.5 xx 10^(5))=2xx 10^(11)`
Promotional Banner

Topper's Solved these Questions

  • AIIMS 2018 PAPER 2

    AIIMS PREVIOUS YEAR PAPERS|Exercise PHYSICS|40 Videos

  • AIIMS 2019 25 MAY MORNING SHIFT

    AIIMS PREVIOUS YEAR PAPERS|Exercise PHYSICS|54 Videos

Similar Questions

Explore conceptually related problems

Electric field due to point charge

(a) Derive a relation for the velocity of electromagnetic wave. (b) An electromagnetic wave is travelling along Z-axis is described by magnetic field B_x =4 xx 10^(-10) sin ( omega t - kx -kz)T Find the maximum electric and magnetic field on an alpha particle moving along Y-axis with a speed of 5 xx 10^6 m/s. Charge on the electron is 1.6 xx 10^(-19) C.

Speed of an electron passing undeviated through a region of cross electric and magnetic fields of magnitude 4 xx 10^(5) V/m and 0.02 Wb//m^(2) respectively in meter per second is

A beam of charged particle enters in a region of magnetic field of 5 × 10^(−3) weber m and electric field of 2.5 × 10 4 V m 1 ,acting perpendicularly. Calculate the speed of particles perpendicular to electric and ,magnetic field , if their path remains unchanged. The given charge on particles and mass are 3.2 × 10 − 19 C and 12 × 10 − 31 kg , respectively.calculate the radius of circular path traced by the charged particle if the electric field is removed .

A velocity filter uses the properties of electric and magnetic field to select particles that are moving with a specifice velocity. Charged particles with varying speeds are directed into the filter as shown. The filter consistt of an electric field E and a magnetic field B, each of constant magnitude, directed perpendicular to each other as shown, The charge particles will experience a force due to electric field given by F = qE. If positively charged particles are used, this force is towards right. The moving particle will also experience a force due to magnetic field given by F = qvB . When forces due to the two fields are of equal magnitude, the net force on the particle will be zero, the particle will pass through centre with its path unaltered. The electric and magnetic fields can be adjusted t choose the specific velocity to be filtered. The efect of gravity can be neglected. The electric and magnetic fields are adjusted to detect particles with positive charge q of certain speed v_(0) . which of the following expressions is equal to this speed?

A beam of charged particle enters in a region of magnetic field of 5 xx 10^(-3) weber m and electric field of 2.5 xx 10^(4) Vm^(1) ,acting perpendicularly. Calculate the speed of particles perpendicular to electric and ,magnetic field , if their path remains unchanged. The given charge on particles and mass are 3.2 xx 10^(-19) C " and " 12 xx 10^(-31) kg , respectively.

Electrons move at right angles to a magnetic field of 1.5xx10^(-2) Tesla with a speed of 6xx10^(7)m//s . If the specific charge of the electron is 1.7xx10^(11)C//kg . The radius of the circular path will be

In a certain region of space a uniform and constant electric field and a magnetic field parallel to each other are present. A proton is fired from a point A in the field with speed v=4xx10^(4)m//s at an angle of alpha with the field direction. The proton reaches a point B in the field where its velocity makes an angle beta with the field direction. If (sinalpha)/(sinbeta)=sqrt(3) . Find the electric potential difference between the points A and B . Take m_(p) (mass of proton) =1.6xx10^(-27)kg and c (magnitude of electronic charge) =1.6xx10^(19)C .

Introduction to Electric Field || Electric Field due to Point Charge || Electric Field due to Rod & Ring || S.H.M due to Electric Field OF Ring

When a particle carrying a charge of 200 mu C moves at an angle 30^(@) to a uniform magnetic field of induction 5 xx 10^(-5) Wb//m^(2) with a speed of 2 xx 10^(5) m//s . The force acting on the particle is

AIIMS PREVIOUS YEAR PAPERS-AIIMS 2019 25 MAY EVENING SHIFT-PHYSICS
  1. In hydrogen atom find magnetic field at center in ground. State if Boh...

    Text Solution

    |

  2. <img src="https://d10lpgp6xz60nq.cloudfront.net/physicsimages/RESAIIMS...

    Text Solution

    |

  3. In a isobaric process the work done by a di-atomic gas is 10 J , the h...

    Text Solution

    |

  4. A capacitor of capacitance 15 nF having dielectirc slab of epsilon(r)...

    Text Solution

    |

  5. An ideal gas initially at pressure 1 bar is being compressed from 30m^...

    Text Solution

    |

  6. Distance between sun and earth is 2 xx 10^(8) km, temperature of sun ...

    Text Solution

    |

  7. Number of visible lines in Balmer series.

    Text Solution

    |

  8. Ratio of electric and magnetic field due to moving point charge if it...

    Text Solution

    |

  9. In toroid magnetic field magnetic field on axis will be radius = 0.5 c...

    Text Solution

    |

  10. The current density is a solid cylindrical wire a radius R, as a funct...

    Text Solution

    |

  11. In maxwell's speed distribution curve, for N(2 ) gas, the average of |...

    Text Solution

    |

  12. N(2) gas is heated from 300 kg temperature to 600 k through an isobari...

    Text Solution

    |

  13. Assertion : In desert area, days get hot fastly and the nights get col...

    Text Solution

    |

  14. Assertion : For communication antennae length should be comparable to ...

    Text Solution

    |

  15. Assertion: Amplitude modulation shows more interference than frequency...

    Text Solution

    |

  16. Assertion: For an element generally N ge Z (N = number of neutrons, Z...

    Text Solution

    |

  17. Assertion : Positive feedback is essential for converting a transistor...

    Text Solution

    |

  18. Assertion : Vibrational degree of freedom of a di-atomic gas molecule ...

    Text Solution

    |

  19. Assertion : NH3 is liquidities more easily than CO(2). Reason : Cri...

    Text Solution

    |

  20. Assertion : Even though net external force on a body is zero, momentum...

    Text Solution

    |