Proton with kinetic energy of 1 MeV moves from south to north. It gets an acceleration of $$10^{12}$$ m/s$$^2$$ by an applied magnetic field (west to east). The value of magnetic field: (Rest mass of proton is $$1.6 \times 10^{-27}$$ kg)

We are given that a proton possesses a kinetic energy of $$1\ \text{MeV}$$ and is travelling from south to north. First, we convert this energy into joules because the usual mechanical formulas employ SI units.

$$1\ \text{MeV}=1\times10^{6}\ \text{eV}=1\times10^{6}\times1.6\times10^{-19}\ \text{J}=1.6\times10^{-13}\ \text{J}$$

The classical relation between kinetic energy and speed is stated as

$$K=\frac12\,m\,v^{2}$$

where $$K$$ is the kinetic energy, $$m$$ the mass and $$v$$ the speed. Substituting $$K=1.6\times10^{-13}\ \text{J}$$ and $$m=1.6\times10^{-27}\ \text{kg}$$ (rest mass of the proton), we get

$$1.6\times10^{-13}=\frac12\,(1.6\times10^{-27})\,v^{2}$$

Multiplying both sides by 2 to remove the fraction,

$$2\times1.6\times10^{-13}=(1.6\times10^{-27})\,v^{2}$$

Notice that the factor $$1.6$$ appears on both sides, so it cancels:

$$2\times10^{-13}=10^{-27}\,v^{2}$$

Dividing by $$10^{-27}$$, we have

$$v^{2}=2\times10^{14}$$

Taking the square root,

$$v=\sqrt{2}\times10^{7}\ \text{m s}^{-1}\approx1.414\times10^{7}\ \text{m s}^{-1}$$

Now, the proton experiences an acceleration of $$10^{12}\ \text{m s}^{-2}$$ due to a magnetic field directed from west to east, while its velocity is south to north. The magnetic force formula is first stated:

$$\mathbf{F}=q\,\mathbf{v}\times\mathbf{B}$$

The magnitude of this force when $$\mathbf{v}$$ and $$\mathbf{B}$$ are perpendicular is

$$F=q\,v\,B$$

According to Newton’s second law, this force produces an acceleration $$a$$, so

$$F=m\,a$$

Equating the two expressions for the force gives

$$m\,a=q\,v\,B$$

Solving for the magnetic field magnitude $$B$$,

$$B=\frac{m\,a}{q\,v}$$

We substitute $$m=1.6\times10^{-27}\ \text{kg}$$, $$a=10^{12}\ \text{m s}^{-2}$$, $$q=1.6\times10^{-19}\ \text{C}$$ and $$v=1.414\times10^{7}\ \text{m s}^{-1}$$:

$$B=\frac{(1.6\times10^{-27})(10^{12})}{(1.6\times10^{-19})(1.414\times10^{7})}$$

Simplifying the numerator,

$$1.6\times10^{-27}\times10^{12}=1.6\times10^{-15}$$

Simplifying the denominator,

$$(1.6\times10^{-19})(1.414\times10^{7})=1.6\times1.414\times10^{-12}=2.2624\times10^{-12}$$

Hence,

$$B=\frac{1.6\times10^{-15}}{2.2624\times10^{-12}}=\frac{1.6}{2.2624}\times10^{-3}\ \text{T}$$

Calculating the numerical ratio,

$$\frac{1.6}{2.2624}\approx0.707$$

Therefore,

$$B\approx0.707\times10^{-3}\ \text{T}=0.707\ \text{mT}$$

This value matches closely with $$0.71\ \text{mT}$$.

Hence, the correct answer is Option A.