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NVAMediumJEE 2023Force on Moving Charge

JEE Physics 2023 Question with Solution

A proton with a kinetic energy of 2.0 eV2.0 \, \text{eV} moves into a region of uniform magnetic field of magnitude π2×10−3 T\dfrac{\pi}{2} \times 10^{-3} \, \text{T}. The angle between the direction of magnetic field and velocity of proton is 60∘60^\circ. The pitch of the helical path taken by the proton is _____ cm.

Answer

Correct answer:40

Step-by-step solution

Standard Method

Given: B=π2×10−3 TB = \frac{\pi}{2} \times 10^{-3} \, \text{T}, K.E.=2.0 eVK.E. = 2.0 \, \text{eV}, m=1.67×10−27 kgm = 1.67 \times 10^{-27} \, \text{kg}, charge on proton =1.6×10−19 C= 1.6 \times 10^{-19} \, \text{C}.

Find: The pitch of the helical path.

From the kinetic energy formula,

K.E.=12mv2K.E. = \frac{1}{2}mv^2

so,

v=2K.E.mv = \sqrt{\frac{2K.E.}{m}}

The pitch of the helical path is given by

Pitch=vcos⁡60∘×time period of one rotation\text{Pitch} = v \cos 60^\circ \times \text{time period of one rotation}

Using the expressions shown,

v=2×2×10−191.6×10−27v = \sqrt{\frac{2 \times 2 \times 10^{-19}}{1.6 \times 10^{-27}}}

and

T=2πmeBT = \frac{2\pi m}{eB}

Therefore,

Pitch=vcos⁡60∘×2πmeB\text{Pitch} = v \cos 60^\circ \times \frac{2\pi m}{eB}

Substituting the given values as shown in the solution,

=2×104×12×4×10−5= 2 \times 10^4 \times \frac{1}{2} \times 4 \times 10^{-5} =4×101 cm= 4 \times 10^1 \, \text{cm}

Therefore, the pitch of the helical path is 40 cm40 \, \text{cm}.

Using parallel velocity and time period

Given: A proton enters a uniform magnetic field with speed making an angle of 60∘60^\circ with the field.

Find: The pitch of the helical path.

In helical motion, the component of velocity parallel to the magnetic field remains unchanged. Hence,

v∥=vcos⁡60∘v_\parallel = v \cos 60^\circ

The particle completes circular motion due to the perpendicular component, and the time period of one revolution is

T=2πmeBT = \frac{2\pi m}{eB}

So the pitch is distance moved along the field in one time period:

Pitch=v∥T=vcos⁡60∘⋅2πmeB\text{Pitch} = v_\parallel T = v \cos 60^\circ \cdot \frac{2\pi m}{eB}

Using the velocity obtained from kinetic energy,

v=2K.E.mv = \sqrt{\frac{2K.E.}{m}}

and then substituting the numerical values given in the solution leads to

Pitch=40 cm\text{Pitch} = 40 \, \text{cm}

Thus, the required numerical value is 40.

Common mistakes

  • Using the full velocity for circular motion and also for pitch without resolving components. This is wrong because only the component parallel to the magnetic field contributes to pitch. Use vcos⁡60∘v \cos 60^\circ for the axial motion.

  • Confusing the time period of circular motion with the pitch directly. Pitch is not the radius or circumference; it is the distance travelled along the magnetic field in one complete revolution, so use Pitch=v∥T\text{Pitch} = v_\parallel T.

  • Leaving the final answer with units in the answer field. For a numerical value answer, units belong in the solution and question only; the answer field should contain only the number.

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