Suppose that a proton travelling in vacuum with velocity `V_(1)` at right angles to a uniform magnetic field experiences twice the force that an `alpha`-particle experiences when it is travelling along into same path with velocity `V_(2)`. The ratio `(V_(1))/(V_(2))` is

A charged particle moves with velocity v in a uniform magnetic field vecB . The magnetic force experienced by the particle is

A proton and an alpha particle projected with same velocity in uniform transverse magnetic field then

A n electrically charged particle enters into a uniform magnetic induction field in a direction perpendicular to the field with a velocity v then it travels

A proton and an alpha- particle enter a uniform magnetic field at right angles to the field with same speed. The ratio of the periods of alpha- particle and proton is

If an alpha- particle moving with velocity 'v' enters perpendicular to a magnetic field then the magnetic force acting on it will be-

A wire moves with a velocity ''v'' through a magnetic field and experiences an induced charge sepration as shown. Then the direction of the magnetic field is

A electron e^(-) is accelerated by V volts experiences a force F, when it enters in a uniform magnetic field. What will the force experienced when it is accelerated by 2V?

An electron is accelerated from rest through a potential difference V. This electron experiences a force F in a uniform magnetic field. On increasing the potential difference to V', the force experienced by the electron in the same magnetic field becomes 2F. Then, the ratio (V'//V) is equal to