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MCQMediumJEE 2025Force on Moving Charge

JEE Physics 2025 Question with Solution

Uniform magnetic fields of different strengths B1B_1 and B2B_2, both normal to the plane of the paper, exist as shown in the figure. A charged particle of mass mm and charge qq, at the interface at an instant, moves into region 22 with velocity vv and returns to the interface. It continues to move into region 11 and finally reaches the interface. What is the displacement of the particle during this movement along the interface?

Two horizontal regions separated by an interface, with crosses indicating uniform magnetic fields into the plane; upper region labeled B1 Region 1 and lower region labeled B2 Region 2.

Consider the velocity of the particle to be normal to the magnetic field and B2>B1B_2 > B_1.

  • A

    mvqB1(1−B2B1)×2\frac{mv}{qB_1} \left( 1 - \frac{B_2}{B_1} \right) \times 2

  • B

    mvqB1(1−B1B2)\frac{mv}{qB_1} \left( 1 - \frac{B_1}{B_2} \right)

  • C

    mvqB1(1−B2B1)\frac{mv}{qB_1} \left( 1 - \frac{B_2}{B_1} \right)

  • D

    mvqB1(1−B1B2)×2\frac{mv}{qB_1} \left( 1 - \frac{B_1}{B_2} \right) \times 2

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: A charged particle with mass mm, charge qq and speed vv moves perpendicular to uniform magnetic fields B1B_1 and B2B_2 in two adjacent regions. Find: the displacement along the interface after the particle goes into region 22, returns to the interface, then goes into region 11 and again reaches the interface.

Since v⃗⊥B⃗\vec{v} \perp \vec{B}, the particle moves in a circular path in each region with radius

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

Therefore, in region 11,

R1=mvqB1R_1 = \frac{mv}{qB_1}

and in region 22,

R2=mvqB2R_2 = \frac{mv}{qB_2}

The motion in each region contributes a diameter along the interface. Hence the net displacement along the interface is

AC=CD−ADAC = CD - AD

Using the diameters in the two regions,

AC=2R1−2R2AC = 2R_1 - 2R_2

Substituting the radii,

AC=2mvqB1−2mvqB2AC = \frac{2mv}{qB_1} - \frac{2mv}{qB_2}

Taking common factor,

AC=2mvqB1(1−B1B2)AC = \frac{2mv}{qB_1}\left(1 - \frac{B_1}{B_2}\right)

Therefore, the displacement along the interface is 2mvqB1(1−B1B2)\frac{2mv}{qB_1}\left(1 - \frac{B_1}{B_2}\right), so the correct option is D.

Using radii in the two magnetic regions

Given: The particle enters two regions having magnetic fields B1B_1 and B2B_2 with B2>B1B_2 > B_1, and its velocity is perpendicular to the magnetic field. Find: the total displacement along the interface.

The magnetic force provides the centripetal force, so the path is circular in each region. The radius depends inversely on magnetic field strength:

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

Thus,

R1=mvqB1,R2=mvqB2R_1 = \frac{mv}{qB_1}, \qquad R_2 = \frac{mv}{qB_2}

Because B2>B1B_2 > B_1,

R2<R1R_2 < R_1

From the geometry of the two semicircular parts of the trajectory, the displacement along the interface equals the difference of the corresponding diameters:

displacement=2R1−2R2\text{displacement} = 2R_1 - 2R_2

Now substitute:

displacement=2(mvqB1−mvqB2)\text{displacement} = 2\left(\frac{mv}{qB_1} - \frac{mv}{qB_2}\right) displacement=2mvq(1B1−1B2)\text{displacement} = \frac{2mv}{q}\left(\frac{1}{B_1} - \frac{1}{B_2}\right)

Rewriting in the option form,

displacement=2mvqB1(1−B1B2)\text{displacement} = \frac{2mv}{qB_1}\left(1 - \frac{B_1}{B_2}\right)

Hence, the correct option is D.

Common mistakes

  • Using R=qBmvR = \frac{qB}{mv} instead of R=mvqBR = \frac{mv}{qB}. This reverses the dependence on magnetic field and gives the wrong displacement. Always obtain the radius from centripetal force balance for magnetic motion.

  • Forgetting that the displacement along the interface involves diameters of the circular arcs, not just radii. Using R1−R2R_1 - R_2 instead of 2R1−2R22R_1 - 2R_2 misses the factor of 22.

  • Assuming the larger magnetic field gives the larger radius. Since R∝1BR \propto \frac{1}{B} for fixed m,q,m, q, and vv, the stronger field B2B_2 produces the smaller circular arc.

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