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JEE Advance - Physics (2017 - Paper 1 Offline - No. 1)

A charged particle (electron or proton) is introduced at the origin (x=0,y=0,z=0) with a given initial velocity $$\overrightarrow v .$$ A uniform electric field $$\overrightarrow E $$ and a uniform magnetic field $$\overrightarrow B $$ exist everywhere. The velocity $$\overrightarrow v ,$$ electric field $$\overrightarrow E $$ and magnetic field $$\overrightarrow B $$ are given in column $$1,2$$ and $$3,$$ respectively. The quantities $${E_0},{B_0}$$ are positive in magnitude.

Column 1 Column 2 Column 3
(I) Electron with $$\overrightarrow v = 2{{{E_0}} \over {{B_0}}}\widehat x$$   (i) $$\overrightarrow E = {E_0}\widehat z$$ (P) $$\overrightarrow B = - {B_0}\widehat x$$
(II) Electron with $$\overrightarrow v = {{{E_0}} \over {{B_0}}}\widehat y$$ (ii) $$\overrightarrow E = - {E_0}\widehat y$$ (Q) $$\overrightarrow B = {B_0}\widehat x$$
(III) Proton with $$\overrightarrow v = 0$$    (iii) $$\overrightarrow E = - {E_0}\widehat x$$ (R) $$\overrightarrow B = {B_0}\widehat y$$
(IV) Proton with $$\overrightarrow v = 2{{{E_0}} \over {{B_0}}}\widehat x$$ (iv) $$\overrightarrow E = {E_0}\widehat x$$ (S) $$\overrightarrow B = {B_0}\widehat z$$
A charged particle (electron or proton) is introduced at the origin (x=0,y=0,z=0) with a given initial velocity $$\overrightarrow v .$$ A uniform electric field $$\overrightarrow E $$ and a uniform magnetic field $$\overrightarrow B $$ exist everywhere. The velocity $$\overrightarrow v ,$$ electric field $$\overrightarrow E $$ and magnetic field $$\overrightarrow B $$ are given in column $$1,2$$ and $$3,$$ respectively. The quantities $${E_0},{B_0}$$ are positive in magnitude.

Column 1 Column 2 Column 3
(I) Electron with $$\overrightarrow v = 2{{{E_0}} \over {{B_0}}}\widehat x$$   (i) $$\overrightarrow E = {E_0}\widehat z$$ (P) $$\overrightarrow B = - {B_0}\widehat x$$
(II) Electron with $$\overrightarrow v = {{{E_0}} \over {{B_0}}}\widehat y$$ (ii) $$\overrightarrow E = - {E_0}\widehat y$$ (Q) $$\overrightarrow B = {B_0}\widehat x$$
(III) Proton with $$\overrightarrow v = 0$$    (iii) $$\overrightarrow E = - {E_0}\widehat x$$ (R) $$\overrightarrow B = {B_0}\widehat y$$
(IV) Proton with $$\overrightarrow v = 2{{{E_0}} \over {{B_0}}}\widehat x$$ (iv) $$\overrightarrow E = {E_0}\widehat x$$ (S) $$\overrightarrow B = {B_0}\widehat z$$
A charged particle (electron or proton) is introduced at the origin (x=0,y=0,z=0) with a given initial velocity $$\overrightarrow v .$$ A uniform electric field $$\overrightarrow E $$ and a uniform magnetic field $$\overrightarrow B $$ exist everywhere. The velocity $$\overrightarrow v ,$$ electric field $$\overrightarrow E $$ and magnetic field $$\overrightarrow B $$ are given in column $$1,2$$ and $$3,$$ respectively. The quantities $${E_0},{B_0}$$ are positive in magnitude.

Column 1 Column 2 Column 3
(I) Electron with $$\overrightarrow v = 2{{{E_0}} \over {{B_0}}}\widehat x$$   (i) $$\overrightarrow E = {E_0}\widehat z$$ (P) $$\overrightarrow B = - {B_0}\widehat x$$
(II) Electron with $$\overrightarrow v = {{{E_0}} \over {{B_0}}}\widehat y$$ (ii) $$\overrightarrow E = - {E_0}\widehat y$$ (Q) $$\overrightarrow B = {B_0}\widehat x$$
(III) Proton with $$\overrightarrow v = 0$$    (iii) $$\overrightarrow E = - {E_0}\widehat x$$ (R) $$\overrightarrow B = {B_0}\widehat y$$
(IV) Proton with $$\overrightarrow v = 2{{{E_0}} \over {{B_0}}}\widehat x$$ (iv) $$\overrightarrow E = {E_0}\widehat x$$ (S) $$\overrightarrow B = {B_0}\widehat z$$
In which case will the particle move in a straight line with constant velocity?
$$\left( {{\rm I}{\rm I}{\rm I}} \right)\left( {ii} \right)\left( R \right)$$
$$\left( {{\rm I}V} \right)\left( i \right)\left( S \right)$$
$$\left( {{\rm I}{\rm I}{\rm I}} \right)\left( {iii} \right)\left( P \right)$$
$$\left( {{\rm I}{\rm I}} \right)\left( {iii} \right)\left( S \right)$$

Explanation

$$\overrightarrow F $$ = q$$\overrightarrow E $$ + q($$\overrightarrow v $$ $$\times$$ $$\overrightarrow B $$) ....... (1)

For a particle to move in a straight line with constant velocity, $$\overrightarrow F $$ = 0. Therefore, from Eq. (1), we get

$$q\overrightarrow E + q(\overrightarrow V \times \overrightarrow B ) = 0$$

$$ \Rightarrow q\overrightarrow E - q(\overrightarrow v \times \overrightarrow B )$$

$$ \Rightarrow \overrightarrow E = - (\overrightarrow v \times \overrightarrow B )$$

For $$\overrightarrow v = {{{E_0}} \over {{B_0}}}\widehat y$$ and $$\overrightarrow B = {B_0}\widehat z$$, we get

$$\overrightarrow v \times \overrightarrow B = \left( {{{{E_0}} \over B}\widehat y} \right) \times ({B_0}\widehat z) = {{{E_0}} \over {{B_0}}} \times {B_0}(\widehat y \times z)$$

Therefore,

$$\overrightarrow v \times \overrightarrow B = {E_0}\widehat x$$ (as y $$\times$$ z = x)

and

$$\overrightarrow E = - {E_0}\widehat x$$

Therefore,

$$v = {{{E_0}} \over {{B_0}}}\widehat y$$; $$\overrightarrow E = - {E_0}\widehat x$$; $$\overrightarrow B = {B_0}\widehat z$$

Hence, option (D) is correct.

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