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Question

(A) The magnetic force $\vec{F}$ on a charged particle moving with velocity $\vec{V}$ in a magnetic field $\vec{B}$ is given by the Lorentz force form

Vedclass · Accepted Answer

$q a \sqrt{b^2 + c^2}$

Vedclass · Answer

(A) The magnetic force $\vec{F}$ on a charged particle moving with velocity $\vec{V}$ in a magnetic field $\vec{B}$ is given by the Lorentz force formula: $\vec{F} = q(\vec{V} 	imes \vec{B})$.Given $\vec{V} = a \hat{i}$ and $\vec{B} = b \hat{j} + c \hat{k}$.Calculating the cross product: $\vec{V} 	imes \vec{B} = (a \hat{i}) 	imes (b \hat{j} + c \hat{k}) = ab(\hat{i} 	imes \hat{j}) + ac(\hat{i} 	imes \hat{k})$.Using unit vector cross products $\hat{i} 	imes \hat{j} = \hat{k}$ and $\hat{i} 	imes \hat{k} = -\hat{j}$,we get: $\vec{V} 	imes \vec{B} = ab \hat{k} - ac \hat{j}$.Therefore,$\vec{F} = q(ab \hat{k} - ac \hat{j})$.The magnitude of the force is $|\vec{F}| = q \sqrt{(ab)^2 + (-ac)^2} = q \sqrt{a^2b^2 + a^2c^2} = qa \sqrt{b^2 + c^2}$.

$A$ particle of charge $q$ moves with a velocity $\vec{V} = a \hat{i}$ in a magnetic field $\vec{B} = b \hat{j} + c \hat{k}$,where $a$,$b$,and $c$ are constants. The magnitude of the force experienced by the particle is:

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