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Question

Proton with kinetic energy of 1MeV moves from south to north. It gets an acceleration of 1012m/s2 by an applied magnetic field (west to east). The value of magnetic field: (Rest mass of proton is 1.6×1027kg)


  1. 0.71mT

  2. 7.1mT

  3. 71mT

  4. 0.071mT

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Solution

The correct option is A

0.71mT


Step 1: Given Data:

Kinetic Energy =1MeV=1×106eV

1eV=1.6×10-19J

So, kinetic energy will be in terms of joule-

1×106eV=1.6×10-13J

Mass of proton mp=1.6×10-27kg

Charge on proton q=1.6×10-19C

Acceleration of particle =1012m/s2

Step 2: Formula Used:

Kinetic Energy=12mv2

Lorentz FormulaF=qvBsinθ

Lorentz FormulaF=qvBsin90=qvB

Where qis an electric charge on the particle

V is the velocity of the particle

B is the magnetic field

ForceF=m×a

From the above two formulas

qvB=ma……………………(1)

Step 3: Calculating the velocity of the proton:

The kinetic energy using the formula,

KE=12mpv21.6×10-13=12×1.6×10-27×v2v2=2×1014v=2×107m/s

The velocity of the proton is v=2×107m/s
Step 4: Calculating the value of the magnetic field:

Using the equation (1)-

The magnetic field can be calculated as-

Bqv=mpaB=mpaqvB=1.6×10-27×10121.6×10-19×2×107=0.71×10-3T=0.71mT

Hence, option (A) is the correct answer.


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