Question

An alpha particle is accelerated by electric field upto a distance of 2m and acquiring kinetic energy 2.5 Mev . Calculate magnetic field
needed to compensate by electric field.​

Answer 1

It has given that an alpha particle is accelerated by electric field upto a distance of 2m and acquiring kinetic energy 2.5 MeV.

we have to calculate magnetic field needed to compensate by electric field.

kinetic energy = workdone by alpha particle

⇒2.5 × 10⁶ eV = magnetic force × 2m

⇒magnetic force = 1.25 × 10⁶ eV/m

we know, magnetic force = BQv

where B is magnetic field , Q is charge and v is velocity of alpha particle.

now , v = √{2K.E/m}

here, K.E = √{2 × 2.5 × 10^6 × 1.6 × 10^-19/4 × 1.67 × 10^-27 }

= √{2/1.67 × 10¹⁴}

= 1.094 × 10^7 m/s

now magnetic force = BQv = 1.25 × 10^6 × eV/m [as Q = 2e ]

⇒B × 2e × 1.094 × 10^7 = 1.25 × 10^6 eV/m

⇒B = (1.25 × 10^6/2.188 × 10^7)

= 0.057 T

Answer 2

Answer:

It has given that an alpha particle is accelerated by electric field upto a distance of 2m and acquiring kinetic energy 2.5 MeV.

we have to calculate magnetic field needed to compensate by electric field.

kinetic energy = workdone by alpha particle

⇒2.5 × 10⁶ eV = magnetic force × 2m

⇒magnetic force = 1.25 × 10⁶ eV/m

we know, magnetic force = BQv

where B is magnetic field , Q is charge and v is velocity of alpha particle.

now , v = √{2K.E/m}

here, K.E = √{2 × 2.5 × 10^6 × 1.6 × 10^-19/4 × 1.67 × 10^-27 }

= √{2/1.67 × 10¹⁴}

= 1.094 × 10^7 m/s

now magnetic force = BQv = 1.25 × 10^6 × eV/m [as Q = 2e ]

⇒B × 2e × 1.094 × 10^7 = 1.25 × 10^6 eV/m

⇒B = (1.25 × 10^6/2.188 × 10^7)

= 0.057 T