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JEE Advanced 2017 · Paper 1, Physics question 15

Passage

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 would the particle move in a straight line along the negative direction of yy-axis (i.e., move along −y^-\hat{y})?

Choose one option
Answer and worked solution

Official answer: B

Option (B) is (III)(ii)(R): a proton at rest, with E⃗=−E0y^\vec E=-E_0\hat y and B⃗=B0y^\vec B=B_0\hat y.

  • The electric force pushes it along −y^-\hat y.
  • Its velocity stays parallel to B⃗\vec B, so v⃗×B⃗=0\vec v\times\vec B=0.

It moves in a straight line along −y^-\hat y.

(B)

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