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The magnetic force acting on a charged particle of charge –2 $\mu$C in a magnetic field of 2 T acting in y direction, when the particle velocity is $(2\hat{i} + 3\hat{j})\times10^6$ m $s^{-1}$, is
A
8 N in z-direction
B
8 N in –z direction
C
4 N in z-direction
D
8 N in y-direction
Detailed Solution
$\vec{F} = q(\vec{v}\times\vec{B})$ with $q = -2\times10^{-6}$ C, $\vec{v} = (2\hat{i} + 3\hat{j})\times10^6$ m/s, $\vec{B} = 2\hat{j}$ T
$\vec{v}\times\vec{B} = (2\hat{i} + 3\hat{j})\times10^6\times2\hat{j} = 4\times10^6(\hat{i}\times\hat{j}) + 6\times10^6(\hat{j}\times\hat{j})$
$\hat{i}\times\hat{j} = \hat{k}$ and $\hat{j}\times\hat{j} = 0$, so $\vec{v}\times\vec{B} = 4\times10^6\hat{k}$
$\vec{F} = -2\times10^{-6}\times4\times10^6\hat{k} = -8\hat{k}$ N
The force is 8 N in the –z direction.
Note: the first two options are printed identically in the source ('8N in z – direction'); the minus sign of one has been lost. They are taken here as +z and –z, with –z as the keyed answer.
$\vec{v}\times\vec{B} = (2\hat{i} + 3\hat{j})\times10^6\times2\hat{j} = 4\times10^6(\hat{i}\times\hat{j}) + 6\times10^6(\hat{j}\times\hat{j})$
$\hat{i}\times\hat{j} = \hat{k}$ and $\hat{j}\times\hat{j} = 0$, so $\vec{v}\times\vec{B} = 4\times10^6\hat{k}$
$\vec{F} = -2\times10^{-6}\times4\times10^6\hat{k} = -8\hat{k}$ N
The force is 8 N in the –z direction.
Note: the first two options are printed identically in the source ('8N in z – direction'); the minus sign of one has been lost. They are taken here as +z and –z, with –z as the keyed answer.