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JEE Advanced 2017 · Paper 1, Physics question 13

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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 move in a straight line with constant velocity?

Choose one option
Answer and worked solution

Official answer: D

Constant velocity needs E⃗+v⃗×B⃗=0\vec E+\vec v\times\vec B=0.

Option (D) is (II)(iii)(S): an electron with v⃗=E0B0y^\vec v=\dfrac{E_0}{B_0}\hat y, E⃗=−E0x^\vec E=-E_0\hat x and B⃗=B0z^\vec B=B_0\hat z.
v⃗×B⃗=E0(y^×z^)=E0x^,E⃗+v⃗×B⃗=0.\vec v\times\vec B=E_0(\hat y\times\hat z)=E_0\hat x,\qquad\vec E+\vec v\times\vec B=0.
The net force is zero (for either sign of charge), so the velocity is constant.

The others fail. For example, in (A) the proton starts at rest and is pushed along −y^-\hat y by E⃗\vec E.

(D)

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