First slide

Force on a charged particle moving in a magnetic field

Question

A particle of charge -16 x10^-18C 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 a. long the negative Z-axis. If the charged particle continues moving along the X-axis, the magnitude of B is

Moderate

A

10³ Wb/m²
B

10⁵ wb/m²
C

10¹⁶ wb / m²
D

10^-3 wb /m²

Solution

The situation is shown in fig .We know that,
F= q (E+ V x B)
or F = qE + q (v x B)
Now Fe = gE
$\begin{aligned} = - 16 \times 10^{- 18} \times 10^{4} (- k) \\ = 16 \times 10^{- 14} k \end{aligned}$
And $F_{m} = - 16 \times 10^{- 18} (10 i \times B j)$
$= - 16 \times 10^{17} B (k)$
$∴ F = F_{e} + F_{m} = 16 \times 10^{- 14} k - 16 \times 10^{- 17} Bk$
Since, the particle will continue to move along +X-axis, so resultant force is zero. Therefore, F_e + F_m = 0
or $16 \times 10^{- 14} k = 16 \times 10^{- 17} Bk$
$∴ B = 10^{3} Wb / m^{2}$

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