A particle of charge \(-16 \times 10^{-18}\) coulomb moving with velocity \(10 \mathrm{~m} \mathrm{~s}^{-1}\) along the \(X\)-axis enters a region
in Physics
21 views
3 Votes
3 Votes

A particle of charge \(-16 \times 10^{-18}\) coulomb moving with velocity \(10 \mathrm{~m} \mathrm{~s}^{-1}\) along the \(X\)-axis enters a region, where a magnetic field of induction \(B\) is along the \(Y\)-axis and an electric field of magnitude \(10^{4} \mathrm{~V}^{-1}\) is along the negative \(Z\)-axis. If the charged particle continues moving along the \(X\)-axis, the magnitude of \(B\) is

(a) \(10^{3} \mathrm{~Wb} \mathrm{~m}^{-2}\)

(b) \(10^{5} \mathrm{~Wb} \mathrm{~m}\)

(c) \(10^{16} \mathrm{~Wb} \mathrm{~m}^{-2}\)

(d) \(10^{-2} \mathrm{~Wb} \mathrm{~m}^{-2}\)

in Physics
by
4.4k Points

1 Answer

2 Votes
2 Votes
 
Best Answer

The correct option of this question will be (a).

Solution —

Given $q=-16 \times 10^{-18} C , v=10 \hat{ m s } s ^{-1}$

$\vec{B} =B \hat{j}, \vec{E}=-10^{4} \hat{k} Vm ^{-1}$

$\vec{F}_{e} =q \vec{E}=-16 \times 10^{-18} \times\left(-10^{4} \hat{k}\right)$

=$16 \times 10^{-14} \hat{k} N$

$\vec{F}_{m} =q(\vec{v} \times \vec{B})=-16 \times 10^{-18}(10 \hat{i} \times B \hat{j})$

=$-16 \times 10^{-17} B \hat{k} N$

As the particle continues to move along the same direction

$F_{m}=F_{e}$

$\therefore \quad 16 \times 10^{-17}$

$B$ = $16 \times 10^{-14}$

$\text { or } B=10^{3} Wb m ^{-2}$

Selected by
by
4.4k Points

Related Questions

1 Vote
1 Vote
0 Answers 24 Views

Welcome To Informesia

Informesia Helps You To Prepare India's All States Board Exams (CBSE, ICSE, UP Board, BSEB, HPBOSE, RBSE, MSBSHSE) and Competitive Exam Like JEE (Main + Advanced), AIIMS, NEET, KVPY, NTSE, BITSAT, Olympiad, CLAT...etc.
For Any Query or Suggestion, Please Contact us on :
[email protected]

Connect us on Social Media