1. Formula for radius of curvature in a magnetic field

When a charged particle enters a uniform magnetic field perpendicular to it, the magnetic force provides the centripetal force:

$qvB = \frac{mv^2}{r}$

So,

$r = \frac{mv}{qB} = \frac{p}{qB}$

where $p = mv$ is the linear momentum.

Thus, for particles entering with equal momentum in the same magnetic field,

$r \propto \frac{1}{|q|}$

The curvature is inversely proportional to radius, so smaller radius means greater curvature.

2. Check Assertion A

Assertion A: If $\mathrm{O}^{2-}$ and $\mathrm{H}^+$ enter normal to the magnetic field with equal momentum, then the path of $\mathrm{O}^{2-}$ ion has a smaller curvature than that of $\mathrm{H}^+$.

For $\mathrm{O}^{2-}$:

$|q| = 2e$

For $\mathrm{H}^+$:

$|q| = e$

Using

$r = \frac{p}{|q|B}$

with equal $p$ and same $B$,

$r_{\mathrm{O}^{2-}} = \frac{p}{2eB}, \qquad r_{\mathrm{H}^+} = \frac{p}{eB}$

Hence,

$r_{\mathrm{O}^{2-}} < r_{\mathrm{H}^+}$

So $\mathrm{O}^{2-}$ has smaller radius, therefore greater curvature, not smaller curvature.

Hence, Assertion A is false.

3. Check Reason R

Reason R: A proton with same linear momentum as an electron will form a path of smaller radius of curvature on entering a uniform magnetic field perpendicularly.

For proton and electron with same momentum in same magnetic field:

• proton charge magnitude $= e$
• electron charge magnitude $= e$

Therefore,

$r = \frac{p}{eB}$

for both.

So both have same radius of curvature (though they bend in opposite directions because charges are opposite).

Thus the statement that proton forms a path of smaller radius is false.

Hence, Reason R is false.

4. Conclusion

• Assertion A: False
• Reason R: False

So the correct logical outcome should be: Both A and R are false.

But none of the given options states this.

5. Comparison with stored answer

Stored correct answer: C

Option C says: A is true but R is false.

This does not match the physics result. Assertion A is false, not true.

Therefore, I disagree with the stored answer.