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× 10- 27 kg ,Charge of the proton =1.6× 10- 19C )

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× 10- 27 kg ,Charge of the proton =1.6× 10- 19C ) (A) 8× 10- 12 (B) 8× 10- 11 (C) 3× 10- 11 (D) 3× 10- 10

Tardigrade Admin · Accepted Answer

Force on moving charge in magnetic field

F = q v B s in θ

But

θ = 9 0^{\circ}

F = q v B

Hence velocity

v = \frac{2 E}{m}

, (E is the kinetic energy of proton)
Putting the values we get

F = qB \frac{2 E}{m} = 1.6 \times 1 0^{- 19} \times 2.5 \frac{2 \times 2 \times 1 0 ^{6} \times 1.6 \times 1 0 ^{- 19}}{1.6 \times 1 0 ^{- 27}} \Rightarrow F = 8 \times 1 0^{- 12} N

Q. A proton is moving perpendicular to a uniform magnetic field of $2.5 T$ with $2 M e V$ kinetic energy. The force on proton (in $N$ ) is (Mass of proton $= 1.6 \times 1 0^{- 27} k g$ ,Charge of the proton $= 1.6 \times 1 0^{- 19} C$ )

$8 \times 1 0^{- 12}$

$8 \times 1 0^{- 11}$

$3 \times 1 0^{- 11}$

$3 \times 1 0^{- 10}$

Q. 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 =1.6×10−27kg ,Charge of the proton =1.6×10−19C )

8×10−12

8×10−11

3×10−11

3×10−10

Force on moving charge in magnetic field F=qvBsinθ But θ=90∘ F=qvB Hence velocity v=m2E​​ , (E is the kinetic energy of proton) Putting the values we get F=qBm2E​​=1.6×10−19×2.51.6×10−272×2×106×1.6×10−19​​⇒F=8×10−12N

Q. A proton is moving perpendicular to a uniform magnetic field of $2.5 T$ with $2 M e V$ kinetic energy. The force on proton (in $N$ ) is (Mass of proton $= 1.6 \times 1 0^{- 27} k g$ ,Charge of the proton $= 1.6 \times 1 0^{- 19} C$ )

$8 \times 1 0^{- 12}$

$8 \times 1 0^{- 11}$

$3 \times 1 0^{- 11}$

$3 \times 1 0^{- 10}$