A particle of charge -16 x 10-18 coulomb moving with velocity 10 ms-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 104 V/m  is along the negative  z-axis . If the charged particle continues moving along the x-axis , the magnitude of B is

  • Option 1)

    103  Wb/m2

  • Option 2)

    105  Wb/m2

  • Option 3)

    1016  Wb/m2

  • Option 4)

    10-3  Wb/m2

 

Answers (1)
V Vakul

As we learnt in

Magnetic field If V(vector), E (vector) and B (vector) are mutually perpendicular -

Fe=Fm

V=\frac{E}{B}

-

 

 

Particle travels along x-axis.Hence v_{y}=v_{z}=0. Field of induction B is along  y-axis. B_{x}=B_{z}=0.

Electric field is along the negative z-axis.  E_{x}=E_{y}=0

\therefore \, \, Net \, force \, on\, particle\, \, \, \vec{F}=q(\vec{E}+\vec{v}\times \vec{B})

Resolve the motion along the three coordinate axis

\therefore \; \; \; a_{x}=\frac{F_{x}}{m}=\frac{q}{m}(E_{x}+v_{y}B_{z}-v_{z}B_{y})

\; \; \; a_{y}=\frac{F_{y}}{m}=\frac{q}{m}(E_{y}+v_{z}B_{x}-v_{x}B_{z})

\; \; \; a_{z}=\frac{F_{z}}{m}=\frac{q}{m}(E_{z}+v_{x}B_{y}-v_{y}B_{x})

Since  E_{x}=E_{y}=0,\; \; v_{y}=v_{z}=0,\; \; B_{x}=B_{z}=0

\therefore \; \; \; a_{x}=a_{y}=0,\; \; a_{z}=\frac{q}{m}(-E_{z}+v_{x}B_{y})

Again \; \; \; a_{z}=0  as the particle traverse through the region undeflected .

\therefore \; \; \; E_{z}=v_{x}B_{y}

or  B_{y}=\frac{E_{z}}{v_{x}}=\frac{10^{4}}{10}=10^{3}\frac{Wb}{m^{2}}

 

 


Option 1)

103  Wb/m2

Correct option

Option 2)

105  Wb/m2

Incorrect option

Option 3)

1016  Wb/m2

Incorrect option

Option 4)

10-3  Wb/m2

Incorrect option

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