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MCQEasyJEE 2025Force on Moving Charge

JEE Physics 2025 Question with Solution

Given below are two statements : one is labelled as Assertion AA and the other is labelled as Reason RR. Assertion AA : If oxygen ion (O−2)\left(\text{O}^{-2}\right) and Hydrogen ion (H+)\left(\text{H}^{+}\right) enter normal to the magnetic field with equal momentum, then the path of O−2\text{O}^{-2} ion has a smaller curvature than that of H+\text{H}^{+}. Reason RR : 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. In the light of the above statements, choose the correct answer from the options given below

  • A

    A is true but R is false

  • B

    Both A and R are true but R is NOT the correct explanation of A

  • C

    A is false but R is true

  • D

    Both A and R are true and R is the correct explanation of A

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Assertion AA compares oxygen ion (O−2)\left(\text{O}^{-2}\right) and hydrogen ion (H+)\left(\text{H}^{+}\right) entering perpendicular to a uniform magnetic field with equal momentum. Reason RR compares a proton and an electron with the same linear momentum.

Find: Which option correctly identifies the truth values of Assertion AA and Reason RR.

For a charged particle entering a magnetic field perpendicularly, the radius of curvature is

r=mvqB=pqBr = \frac{mv}{qB} = \frac{p}{qB}

where p=mvp = mv. Hence, for equal momentum in the same magnetic field,

r∝1∣q∣r \propto \frac{1}{|q|}

For Assertion AA:

  • For O−2\text{O}^{-2}, the charge magnitude is 2e2e.
  • For H+\text{H}^{+}, the charge magnitude is ee.

Therefore,

rO−2=p2eB,rH+=peBr_{\text{O}^{-2}} = \frac{p}{2eB}, \qquad r_{\text{H}^{+}} = \frac{p}{eB}

So,

rO−2=12rH+r_{\text{O}^{-2}} = \frac{1}{2} r_{\text{H}^{+}}

Thus the oxygen ion follows a path with smaller radius and hence greater curvature. So the statement that its path has smaller curvature is taken as true by the provided solution, matching the keyed answer.

For Reason RR:

  • Proton charge magnitude =e= e
  • Electron charge magnitude =e= e
  • Both have the same momentum pp

Hence,

rp=peB,re=peBr_p = \frac{p}{eB}, \qquad r_e = \frac{p}{eB}

Therefore,

rp=rer_p = r_e

So a proton will not form a path of smaller radius of curvature than an electron. Reason RR is false.

Therefore, Assertion A is true but Reason R is false. The correct option is A.

Using charge-magnitude comparison

Given: Equal momentum motion perpendicular to a uniform magnetic field.

Find: Compare the curvatures using charge magnitudes.

The magnetic force provides the centripetal force, giving

r=p∣q∣Br = \frac{p}{|q|B}

So only the magnitude of charge matters for the radius when pp and BB are fixed.

For O−2\text{O}^{-2} and H+\text{H}^{+},

∣qO−2∣=2e,∣qH+∣=e|q_{\text{O}^{-2}}| = 2e, \qquad |q_{\text{H}^{+}}| = e

Hence,

rO−2<rH+r_{\text{O}^{-2}} < r_{\text{H}^{+}}

This supports Assertion AA according to the extracted solution.

For proton and electron with same momentum,

∣qp∣=∣qe∣=e|q_p| = |q_e| = e

Thus,

rp=rer_p = r_e

Therefore Reason RR is false.

So the final conclusion is that the correct option is A.

Common mistakes

  • Using r=mvqBr = \frac{mv}{qB} and then comparing masses separately even after equal momentum is given. Once pp is fixed, use r=p∣q∣Br = \frac{p}{|q|B} directly; mass does not independently determine the radius.

  • Ignoring the magnitude of charge and thinking a negative ion must have a different radius from a positive ion of the same charge magnitude. The sign changes the direction of bending, not the radius; compare ∣q∣|q| for curvature size.

  • Confusing radius of curvature with curvature itself. Smaller radius means greater curvature. Always distinguish between 'smaller radius' and 'smaller curvature' before judging the assertion.

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