Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

A particle mass m and charge q is in an electric and magnetic field is given by $\overrightarrow{E} = 2\widehat{i}+3\widehat{j}; \overrightarrow{B} = 4\widehat{j}+6\widehat{k}$ . The charged particle is shifted from the origin to the point P(x = 1, y = 1)along a straight path. The magnitude of the total work done is:

Option 1)
(0.15)q
Option 2)
(0.35)q

Option 3)
5q
Option 4)
(2.5)q

Answers (1)

best_answer

Force on a charged particle in magnetic field -

$\underset{F}{\rightarrow}=q (\underset{V}{\rightarrow}\times \underset{B}{\rightarrow})$

$F=qVB\sin \theta$

- wherein

$\underset{V}{\rightarrow}$ - velocity of the particle

$\underset{B}{\rightarrow}$ magnetic field

$E=2\hat{i}+3\hat{j}$

$\vec{B}=4\hat{j}++6\hat{k}$

$0(0,0)\; \; \; \; \; \; \; \; \; ,\; \; \; \; \; \; \; \; \; \; \; P(1,1)$

$OP=S=1\hat{i}+1\hat{j}$

$W=F_{net}\cdot S$

$\vec{F_{net}}=9\vec{E}+9\left (\vec{V}\times \vec{B} \right )$

$\vec{V}\times \vec{ B}$ is perpendicular to $\vec{V}$

So $\vec{V}\times \vec{ B}$ is perpendicular to $\vec{S}$

So $W_{B}=9\left ( \vec{V}\times \vec{B} \right )\cdot \vec{S}=0$

$W_{E}=\left ( 9\vec{E} \right )\cdot \vec{S}$

$=\left ( 29\hat{i}+39\hat{j} \right )\left ( 1\hat{i} +1\hat{j}\right )$

$=29+39$

$W_{net}=W_{E}+W_{B}=59$

Option 1)

(0.15)q

Option 2)

(0.35)q

Option 3)
5q
Option 4)

(2.5)q

Posted by

admin

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2024 Topper: Farmer’s son Gajare Nilkrishna says ‘keep questioning until you understand’

anam-khan

NIT Puducherry Cut-off 2024: Computer Science and Engineering had highest closing rank last year

anam-khan

What after JEE Main results 2024? All you need to know about JEE Adv, CSAB, JoSAA counselling

Latest Question