First slide

Force on a charged particle moving in a magnetic field

Question

If a particle of charge $10^{- 12} C$ moving along the x-direction with a velocity 10⁵ m/s experiences a force of $10^{- 10}$ newton in y-direction due to a magnetic field, then the minimum magnitude of magnetic field is

Easy

A

$10^{- 3} T i n z d i r e c t i o n$
B

$10^{- 15} T in z - direction$
C

$6 .25 \times 10^{- 3} T in z - direction$
D

$10^{- 3} T$ in negative z direction

Solution

$\begin{array}{l} F = qvB \sin θ \Rightarrow B = \frac{F}{qv \sin θ} \\ B_{\min} = \frac{F}{qv} (when θ = 90^{0}) \end{array}$
$∴ B_{\min} = \frac{F}{qv} = \frac{10^{- 10}}{10^{- 12} \times 10^{5}} = 10^{- 3} T$ in negative z direction .

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