Question

A proton is moving perpendicular to a uniform magnetic field of $2.5T$ with $2MeV$ kinetic energy. The force on proton (in $N$ ) is (Mass of proton $=1.6\times 10^{- 27} \, kg$ ,Charge of the proton $=1.6\times 10^{- 19}C$ )

• A. $8\times 10^{- 12}$
• B. $8\times 10^{- 11}$
• C. $3\times 10^{- 11}$
• D. $3\times 10^{- 10}$

Answer 1

Answer: (A)

Force on moving charge in magnetic field
$F=qvBsin \theta $
But $\theta =90^\circ $
$F=qvB$
Hence velocity $v= \, \sqrt{\frac{2 E}{m}}$ , (E is the kinetic energy of proton)
Putting the values we get
$F=qB\sqrt{\frac{2 E}{m}}=1.6\times 10^{- 19}\times 2.5\sqrt{\frac{2 \times 2 \times 10^{6} \times 1 . 6 \times 10^{- 19}}{1 . 6 \times 10^{- 27}}}\Rightarrow F=8\times 10^{- 12}N$