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  • A particle of charge 16×10-16 C moving with velocity 10ms-1 along x− axis enters enters a region where magnetic field of induction&nb

A particle of charge 16×10-16 C moving with velocity 10ms-1 along x− axis enters enters a region where magnetic field of induction \overrightarrow{B} is along the y-axis and an electric field of magnitude 104V-1is along the negative z-axis. If the charged particle continues moving along x−axis, the magnitude of magnitude  of \overrightarrow{B} is :

Option: 1

16×103Wbm-2

   


Option: 2

2×103Wbm-2


Option: 3

1×103Wbm-2


Option: 4

4×103Wbm-2


Answers (1)

best_answer

\text {Given that, Particle of charge}\Rightarrow $q=1.6 \times 10^{-19} \mathrm{C}$\\ \ Velocity \ \ \Rightarrow $v=10 \mathrm{m} / \mathrm{s}$\\ \text {Field induction}\Rightarrow $B=10^{4} V / m$

Now, particle travel along x-axis

Then, V_y=V_z=0

And field induction B is along y-axis

Then,B_x=B_z=0

 

The electric field is along the negative z-axis

Then,  E_x=E_y=0

Now, net force on particle is

F=q(\vec{E}+\vec{v} \times \vec{B})

Resolve the motion along the three coordinates 

\begin{aligned} &F_{x}=m a_{x}\\ &a_{x}=\frac{F_{x}}{m}=\frac{q}{m}\left(E_{x}+v_{y} B_{z}-v_{z} B_{y}\right)\\ &a_{y}=\frac{q}{m}\left(E_{y}+v_{z} B_{x}-v_{x} B_{z}\right)\\ &a_{z}=\frac{q}{m}\left(E_{z}+v_{x} B_{y}-v_{x} B_{y}\right) \end{aligned}

\text { So, } E_{x}=E_{y}=0, v_{y}=v_{z}=0, B_{x}=B_{z}=0

So \ we \ get \\ a_{x}=a_{y}=0\\ a_{z}=\frac{q}{m}\left(-E_{z}+v_{x} B_{y}\right)

Now, again

 a_z  as the particle traverse through the region un-deflected

\begin{aligned} &E_{z}=v_{x} B_{y}\\ &B_{y}=\frac{E_{z}}{v_{x}}\\ &B_{y}=\frac{10^{4}}{10}\\ &B_{y}=10^{3} w b / m^{2} \end{aligned}

Hence, the magnitude of B is  10^{3} w b / m^{2}

 

 

Posted by

Ritika Jonwal

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