Question

A particle of charge -16 x 10^-18 C moving with velocity 10 m/s along the X-axis enters a region where a magnetic field of induction B is along Y-axis and 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 ____________.

Options

• 10³ Wb/m²

• 10⁵ Wb/m²

• 10⁶ Wb/m²

• 10⁷ Wb/m²

Answer 1

A particle of charge -16 x 10^-18 C moving with velocity 10 m/s along the X-axis enters a region where a magnetic field of induction B is along Y-axis and 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 10³ Wb/m².

Explanation:

As per the figure,

F = q(E + v x B) = qE + q(v x B)

Now, F_e= qE

`=-12 xx 10^-18 xx 10^4 (- hat"k")`

`= 12 xx 10^-14  hat "k"`

`"And F"_"m" = -12 xx 10^-18 (10  hat"i" xx "B"  hat"j")`

`= -12 xx 10^-17 "B" (hat"k")`

 F = Fe + Fm

`= 12 xx 10^-14  hat"k" - 12 xx 10^-17  "B"  hat"k"`

Since, the particle will continue to move along +X-axis, so resultant force is zero.

Therefore, F_e + F_m = 0

`therefore 12 xx 10^-14   hat"k"`

` = 12 xx 10^-17  "B"  hat"k"`

`therefore "B" = 10^3  "Wb/m"^2`