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JEE Advanced 2017 · Paper 1, Physics question 14

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Answer Q.13, Q.14 and Q.15 by appropriately matching the information given in the three columns of the following table.

A charged particle (electron or proton) is introduced at the origin (x=0,y=0,z=0)(x = 0, y = 0, z = 0) with a given initial velocity v⃗\vec{v}. A uniform electric field E⃗\vec{E} and a uniform magnetic field B⃗\vec{B} exist everywhere. The velocity v⃗\vec{v}, electric field E⃗\vec{E} and magnetic field B⃗\vec{B} are given in columns 1, 2 and 3, respectively. The quantities E0E_0, B0B_0 are positive in magnitude.

Column 1 Column 2 Column 3
(I) Electron with v⃗=2E0B0x^\vec{v} = 2\frac{E_0}{B_0}\hat{x} (i) E⃗=E0z^\vec{E} = E_0\hat{z} (P) B⃗=−B0x^\vec{B} = -B_0\hat{x}
(II) Electron with v⃗=E0B0y^\vec{v} = \frac{E_0}{B_0}\hat{y} (ii) E⃗=−E0y^\vec{E} = -E_0\hat{y} (Q) B⃗=B0x^\vec{B} = B_0\hat{x}
(III) Proton with v⃗=0\vec{v} = 0 (iii) E⃗=−E0x^\vec{E} = -E_0\hat{x} (R) B⃗=B0y^\vec{B} = B_0\hat{y}
(IV) Proton with v⃗=2E0B0x^\vec{v} = 2\frac{E_0}{B_0}\hat{x} (iv) E⃗=E0x^\vec{E} = E_0\hat{x} (S) B⃗=B0z^\vec{B} = B_0\hat{z}

In which case will the particle describe a helical path with axis along the positive zz direction?

Choose one option
Answer and worked solution

Official answer: A

Option (A) is (IV)(i)(S): a proton with v⃗=2E0B0x^\vec v=2\dfrac{E_0}{B_0}\hat x, E⃗=E0z^\vec E=E_0\hat z and B⃗=B0z^\vec B=B_0\hat z.

  • v⃗⊥B⃗\vec v\perp\vec B, so the proton circles in the xyxy-plane.
  • E⃗\vec E along +z^+\hat z accelerates it steadily along +z+z, which is parallel to B⃗\vec B and unaffected by it.

Together these give a helix with its axis along +z+z (with increasing pitch).

(A)

Topics:Motion in magnetic and electric fieldsForce on a moving charge
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