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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)

As we learnt in

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

F_e=F_m

-

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

Posted by

Vakul

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