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A 2 MeV proton is moving perpendicular to a uniform magnetic field of 2.5 Tesla. The force on the proton is
A). $\text{2}\text{.5 }\times \text{ 1}{{\text{0}}^{-\text{10}}}\text{ N}$
B). $\text{8 }\times \text{ 1}{{\text{0}}^{-\text{11}}}\text{ N}$
C). $\text{2}\text{.5 }\times \text{ 1}{{\text{0}}^{-\text{11}}}\text{ N}$
D). $\text{8 }\times \text{ 1}{{\text{0}}^{-\text{12}}}\text{ N}$

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Last updated date: 13th Jun 2024
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Answer
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Hint: By using the formula of force acting on the particle and its kinetic energy, we will find out the relation for the force with the given values, and then upon calculation, we will get the desired result.

Formula used:
F = qvB
$K_e =\dfrac{1}{2}mv^2$

Complete step by step answer:
The given situation can be illustrated as follows:
seo images


Now, from the given question, we can find the force on the proton using the conservation of energy.
We know that the mass of a proton is
${{\text{M}}_{\text{p}}}\text{ = 1}\text{.6 }\times \text{ 1}{{\text{0}}^{-27}}\text{ kg}$
And the charge in a single proton is
$e\text{ = 1}\text{.6 }\times \text{ 1}{{\text{0}}^{-19}}\text{ C}$
Further,
From energy conservation theory, we know that the formula of Force is
$\text{F = qvB }\ldots \text{(i)}$
Where,
     F = force
     q = electric charge
     v = velocity
     B = magnetic field
We also know,
The formula for kinetic energy is
$\dfrac{\text{m}{{\text{v}}^{2}}}{\text{2}}\text{=}{{\text{K}}_{\text{e}}}\text{ }\ldots \text{(ii)}$
Where,
     ${{\text{K}}_{\text{e}}}$ = kinetic energy
     v = velocity
     m = mass of body
Equating both equations (i) and (ii) with respect to v, we get,
$\text{v = }\sqrt{\dfrac{2{{\text{K}}_{\text{e}}}}{\text{m}}}$
Now,
Since we know the value of v,
$\therefore \text{ f = q }\times \text{ }\sqrt{\dfrac{2{{\text{K}}_{\text{e}}}}{\text{m}}}\text{ }\times \text{ B}$
By equating this we get,
$\text{f = 1}\text{.6 }\times \text{ 1}{{\text{0}}^{-\text{19}}}\text{ }\times \text{ }\sqrt{\dfrac{2\text{ }\times \text{ 2 }\times \text{ 1}{{\text{0}}^{\text{6}}}\text{ }\times \text{ 1}\text{.6 }\times \text{ 1}{{\text{0}}^{-19}}}{\text{1}\text{.6 }\times \text{ 1}{{\text{0}}^{-\text{27}}}}}\text{ }\times \text{ 2}\text{.5}$
$=\text{ 8 }\times \text{ 1}{{\text{0}}^{-\text{12}}}\text{ N}$
Therefore, the correct option is Option D.

Note: Since the charge on proton inside an atom is always positive, and that of an electron is always negative, the overall charge of an atom is neutral. This happens because of the same number of protons and the electrons cancel each other. It should also be known to us that the charge on the proton is because of the quarks that make up the nucleons. By the nucleons, we mean the protons and the neutrons both taken into consideration at the same time.