1. Interpret the question

The word photon is clearly inconsistent here, because a photon has zero rest mass and cannot be treated with magnetic force as in this problem. Since the rest mass of proton is given, the intended particle is a proton.

So we solve for a proton of kinetic energy $1\,\text{MeV}$ moving in a magnetic field.

2. Use magnetic force as centripetal force

For a charged particle moving perpendicular to a magnetic field,

$F = qvB$

and this provides the acceleration,

$F = ma$

So,

$qvB = ma \quad \Rightarrow \quad B = \frac{ma}{qv}$

We need $v$ from the given kinetic energy.

3. Find speed from kinetic energy

Given kinetic energy:

$K = 1\,\text{MeV} = 10^6 \times 1.6 \times 10^{-19} = 1.6 \times 10^{-13}\,\text{J}$

Using

$K = \frac{1}{2}mv^2$

we get

$v = \sqrt{\frac{2K}{m}}$

Given proton mass:

$m = 1.6 \times 10^{-27}\,\text{kg}$

So,

$v = \sqrt{\frac{2(1.6 \times 10^{-13})}{1.6 \times 10^{-27}}}$

$v = \sqrt{2 \times 10^{14}}$

$v \approx 1.414 \times 10^7\,\text{m/s}$

4. Now calculate magnetic field

Given:

• $a = 10^{12}\,\text{m/s}^2$
• $q = 1.6 \times 10^{-19}\,\text{C}$
• $m = 1.6 \times 10^{-27}\,\text{kg}$
• $v \approx 1.414 \times 10^7\,\text{m/s}$

Thus,

$B = \frac{ma}{qv}$

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

Cancel $1.6$:

$B = \frac{10^{-15}}{1.414 \times 10^{-12}}$

$B \approx 0.707 \times 10^{-3}\,\text{T}$

$B = 7.07 \times 10^{-4}\,\text{T}$

$B = 0.707\,\text{mT}$

So,

$\boxed{B \approx 0.71\,\text{mT}}$

5. Check direction consistency

Velocity is from south to north, and magnetic field is from west to east. These are perpendicular, so magnetic force can indeed produce the required acceleration. The magnitude calculation is valid.

6. Option matching

$0.71\,\text{mT}$ corresponds to:

$\boxed{\text{Option A}}$

7. Comparison with stored correct answer

Stored correct answer: A

Derived answer: A

They agree.