First slide

Magnetic force

Question

A particle with charge -5.60 nC is moving in a uniform magnetic field $\vec{B} = - (1.25 T) \hat{k}$ . The magnetic force on the particle is measured to be $\vec{F} = - (3.36 \times 10^{- 7} N) \hat{i} + (7.42 \times 10^{- 7} N) \hat{j}$ . Calculate one of the components of velocity of the particle from this information.

Moderate

A

$- 48 {ms}^{-1}$
B

$4.8 {ms}^{-1}$
C

$40.8 {ms}^{-1}$
D

$- 40.8 {ms}^{-1}$

Solution

Let velocity of the particle is $\vec{v} = v_{x} \hat{i} + v_{y} \hat{j} + v_{z} \hat{k}$
Then $\vec{F} = q \vec{v} \times \vec{B}$
$\Rightarrow - 3.36 \times 10^{- 7} \hat{i} + 7.42 \times 10^{- 7} \hat{j}$ $= - 5.6 \times 10^{- 9} (v_{x} \hat{i} + v_{y} \hat{j} + v_{z} \hat{k}) \times (- 1.25) \hat{k}$
$\Rightarrow - 3.36 \hat{i} + 7.42 \hat{j} = 7 \times 10^{- 2} (- v_{x} \hat{j} + v_{y} \hat{i})$
$\Rightarrow v_{x} = - \frac{7.42 \times 10^{2}}{7} = - 106 m s^{- 1}$
And $\Rightarrow v_{y} = - \frac{3.36 \times 10^{2}}{7} = - 48 m s^{- 1}$ .

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