The following MCQ appeared in Kerala Medical Entrance 2007 question paper:

**A proton with energy of 2 MeV enters a uniform magnetic field of 2.5 T normally. The magnetic force on the proton is (Take mass of proton to be 1.6****×10 ^{–27} kg)**

**(a) 3****×10 ^{–12 }N **

**(b) 8**

**×10**

^{–10 }N**(c) 8**

**×10**

^{–12 }N**(d) 2**

**×10**

^{–10 }N**(e) 3**

**×10**

^{–10 }N

The velocity ‘v’ of the proton is to be found first using the expression for kinetic energy E (in joule). Note that E = 2 MeV = 2×10^{6}×1.6×10^{–19} joule.

We have E = ½ mv^{2} from which v = √(2E/m) = [2×2×10^{6}×1.6×10^{–}^{19}/(1.6×10^{–27})]^{1/2 }= 2×10^{7} ms^{–1}. ^{}

Substituting this value of ‘v’ in the expression for magnetic force (F = qvB), we obtain

F = 1.6×10^{–19}×2×10^{7}×2.5 = 8×10^{–12 }N.** **

Note: In the above question we did not take the ** relativistic increase of mass** of the proton into consideration. Since the energy is 2 MeV only and the proton is is fairly heavy, the relativistic increase of mass will be about 0.2% only and you will obtain the velocity of the proton of 2 MeV energy as 1.995×10

^{7}ms

^{–1}. So, the magnetic force will be slightly

*reduced*.

You should be aware of the relativistic increase in mass when you deal with questions like the above one, especially if you are preparing for GRE Physics Exam or AP Physics Exam in which you can expect questions of the type given below:

**An electron with energy of 2 MeV enters a uniform magnetic field of 2.5 T normally. The magnetic force on the electron is nearly (Take the rest mass of electron to be 9.1****×10 ^{–31} kg)**

**(a) 1.17****×10 ^{–10 }N **

**(b) 8**

**×10**

^{–10 }N**(c) 3.35**

**×10**

^{–10 }N**(d) 8**

**×10**

^{–12 }N**(e) 3.14**

**×10**

^{–14 }N

If you calculate the velocity of the electron without considering the relativistic increase in mass, as we did in the previous question, you will get (nearly) v = 8.39×10^{8} ms^{–1} and the magnetic force F = 3.35×10^{–10 }N, nearly. But, the velocity is more than the velocity of light in free space and therefore is absurd. So, (c) is *not* the correct option.

If m_{0 }is the rest mass of the electron and_{ }‘m’ is its mass while moving with kinetic energy E (= 2 MeV), we have

(m – m_{0})c^{2} = E, from which m = m_{0 }+ E/c^{2}. The energy E is to be substituted in joule in this equation

Therefore, m = 9.1×10^{–31 }+ (2×10^{6} ×1.6×10^{–19})/(3 ×10^{8})^{2} = 9.1×10^{–31} +3.55×10^{–30} = 4.46×10^{–30} kg. Note that the mass of the electron has become nearly *five* times its rest mass.

But m = m_{0}/√(1^{ }– v^{2}/c^{2}) so that v = c√[1^{ }– (m_{0}/m)^{2}] = 3×10^{8}×√[1^{ }– (9.1×10^{–31 }/4.46×10^{–30})^{2}] = 2.93×10^{8}** **ms^{–1}

The magnetic force on the electron is qvB = 1.6×10^{–19}×2.93×10^{8}×2.5 = 1.17×10^{–10 }N.** **** **

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