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MCQEasyJEE 2024Force on Moving Charge

JEE Physics 2024 Question with Solution

Two particles X and Y having equal charges are being accelerated through the same potential difference. Thereafter, they enter normally in a region of uniform magnetic field and describe circular paths of radii R1R_1 and R2R_2, respectively. The mass ratio of X and Y is:

  • A

    (R2/R1)2(R_2/R_1)^2

  • B

    (R1/R2)2(R_1/R_2)^2

  • C

    R1/R2R_1/R_2

  • D

    R2/R1R_2/R_1

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: The two particles have equal charge qq, are accelerated through the same potential difference VV, and move normally into a uniform magnetic field BB. Their circular path radii are R1R_1 for particle X and R2R_2 for particle Y.

Find: The mass ratio mXmY\frac{m_X}{m_Y}.

For motion perpendicular to a magnetic field, the radius of the circular path is

R=mvqBR = \frac{mv}{qB}

Since each particle is accelerated through the same potential difference, the gained kinetic energy is

12mv2=qV\frac{1}{2}mv^2 = qV

So,

v=2qVmv = \sqrt{\frac{2qV}{m}}

Substituting this value of vv into the radius formula,

R=mqB2qVmR = \frac{m}{qB}\sqrt{\frac{2qV}{m}}

Thus, for fixed qq, VV, and BB, we get R∝mR \propto \sqrt{m}.

Therefore,

R1R2=mXmY\frac{R_1}{R_2} = \sqrt{\frac{m_X}{m_Y}}

Squaring both sides,

mXmY=(R1R2)2\frac{m_X}{m_Y} = \left(\frac{R_1}{R_2}\right)^2

Therefore, the mass ratio is (R1R2)2\left(\frac{R_1}{R_2}\right)^2 and the correct option is B.

Expanded Derivation

Given: R=mvqBR = \frac{mv}{qB} and 12mv2=qV\frac{1}{2}mv^2 = qV for both particles.

Find: mXmY\frac{m_X}{m_Y} in terms of R1R_1 and R2R_2.

From energy gain,

v=2qVmv = \sqrt{\frac{2qV}{m}}

Now substitute into the magnetic radius expression:

R=mvqB=mqB2qVmR = \frac{mv}{qB} = \frac{m}{qB}\sqrt{\frac{2qV}{m}}

This simplifies to

R=2VmqB2R = \sqrt{\frac{2Vm}{qB^2}}

Hence for the two particles,

R12=2VmXqB2,R22=2VmYqB2R_1^2 = \frac{2Vm_X}{qB^2}, \qquad R_2^2 = \frac{2Vm_Y}{qB^2}

Dividing the two equations,

R12R22=mXmY\frac{R_1^2}{R_2^2} = \frac{m_X}{m_Y}

So,

mXmY=(R1R2)2\frac{m_X}{m_Y} = \left(\frac{R_1}{R_2}\right)^2

Thus, the correct option is B.

Common mistakes

  • Using R∝mR \propto m instead of R∝mR \propto \sqrt{m} is incorrect because the velocity depends on mass after acceleration through the same potential difference. First use 12mv2=qV\frac{1}{2}mv^2 = qV, then substitute into the radius formula.

  • Comparing the radii directly as mXmY=R1R2\frac{m_X}{m_Y} = \frac{R_1}{R_2} is wrong because the proportionality is with the square root of mass. You must square the ratio of radii to obtain the mass ratio.

  • Ignoring that both particles have the same charge and are accelerated through the same potential difference leads to unnecessary extra variables. Since qq, VV, and BB are common, they cancel in the ratio.

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