CCommons
Search✍️ WriteSign in
⚛️ NEET Physics❓ Question

Magnetic force does no work, yet the velocity direction changes. And what's the pitch of the helical path?

Riya Singh12th Sample

10 Oct 2026 · 1 min read

Chapter: Moving Charges and Magnetism. I'm trying to see the story in the equations, but a few lines feel like they contradict each other.

The force on a charge is F⃗=q(v⃗×B⃗)\vec F = q(\vec v\times\vec B). It's always perpendicular to v⃗\vec v, so it does no work and the kinetic energy doesn't change. Fine. But if there's a force there's an acceleration, so the velocity changes. If the velocity changes the KE should change, no? That is the sentence I wrote in my answer and it got a red cross, but it sounds logical to me.

Next: a charged particle enters a uniform field with velocity at angle θ\theta to B⃗\vec B. NCERT says the path is a helix. I can derive the radius r=mvsin⁡θqBr = \dfrac{mv\sin\theta}{qB} from circular motion in the perpendicular plane. But I don't get what the pitch is and why the time period doesn't depend on speed.

Also, for an electron and a proton with the same kinetic energy entering the same field perpendicular to it, which has the larger radius?

WhatsAppPost on X

6 answers & replies

Sign in to join the conversation.

  • [+]Arjun Nair12thSample· 9d ago

    Velocity is a vector. A force perpendicular to v⃗\vec v changes the direction of v⃗\vec v, not its magnitude. KE depends on v2v^2 (speed), not on direction. So the speed stays constant, KE constant, but the velocity vector rotates. Work done is F⃗⋅dr⃗=0\vec F\cdot d\vec r = 0 since F⃗⊥dr⃗\vec F\perp d\vec r. So the force changes momentum (direction) but not energy. That's why a magnetic field can steer a particle but cannot speed it up.

    • [+]Riya Singh12thSample· 9d agoAuthor

      Speed vs velocity again, as always. "A force perpendicular to v changes direction not speed". Writing it on a card. Thanks everyone, this one has a good ending after all.

  • [+]Neel Chatterjee12thSample· 9d ago

    Resolve v⃗\vec v into v∥=vcos⁡θv_\parallel = v\cos\theta along B⃗\vec B and v⊥=vsin⁡θv_\perp = v\sin\theta perpendicular to B⃗\vec B. The parallel part feels no force, so the particle keeps drifting along B⃗\vec B at vcos⁡θv\cos\theta. The perpendicular part goes round in a circle of radius r=mvsin⁡θqBr = \dfrac{mv\sin\theta}{qB}. Circle plus steady drift = helix.

    Time for one circle: T=2πrvsin⁡θ=2πmqBT = \dfrac{2\pi r}{v\sin\theta} = \dfrac{2\pi m}{qB}. The vv cancels, which is why the period (and frequency qB2πm\dfrac{qB}{2\pi m}, called cyclotron frequency) doesn't depend on speed. Pitch = distance along B in one period: p=vcos⁡θ⋅T=2πmvcos⁡θqBp = v\cos\theta\cdot T = \frac{2\pi m v\cos\theta}{qB}

  • [+]Meera Patel11thSample· 9d ago

    Same KE, same field, same charge magnitude: r=mvqB=2mKqBr = \dfrac{mv}{qB} = \dfrac{\sqrt{2mK}}{qB}. So r∝mr\propto\sqrt{m} for a given KK, qq and BB. Proton is about 1836 times heavier so its radius is about 43 times larger (1836≈42.8\sqrt{1836}\approx 42.8). Electron has the smaller radius. They also circulate in opposite senses since the charges are opposite.

  • [+]Dev Agarwal12thSample· 9d ago

    If speed is the same then electron radius is smaller too because it's lighter, and rotation is clockwise for the proton and anticlockwise for the electron when B points out of the page... I'm blanking on the direction, which one is it?

    • [+]Harsh Gupta12thSample· 9d ago

      Use the formula, bro. Take B⃗\vec B out of the page (+z) and a positive charge moving in +x. v⃗×B⃗=x^×z^=−y^\vec v\times\vec B = \hat{x}\times\hat{z} = -\hat{y}. Force on the positive charge is in −y-y, so it turns toward the right-hand side of its motion, i.e. clockwise as you look at the page. A negative charge feels the opposite force, so it turns anticlockwise. I always do this one cross product rather than trying to remember.

More from Riya Singh

Keep reading