If a particle of charge 10^{-12} , C moving a | vedclass

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(D) The magnetic force on a moving charge is given by $\vec{F} = q(\vec{v} 	imes \vec{B})$.Given: $q = 10^{-12} \, C$,$\vec{v} = 10^5 \hat{i} \, m/s$

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$10^{-3} \, T$ in $\hat{z}$ direction

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(D) The magnetic force on a moving charge is given by $\vec{F} = q(\vec{v} 	imes \vec{B})$.Given: $q = 10^{-12} \, C$,$\vec{v} = 10^5 \hat{i} \, m/s$,and $\vec{F} = 10^{-10} \hat{j} \, N$.The magnitude of the force is $F = qvB \sin 	heta$.For the minimum magnetic field,we assume $\sin 	heta = 1$ (i.e.,$	heta = 90^\circ$),so $F = qvB$.Thus,$B_{\min} = \frac{F}{qv} = \frac{10^{-10}}{10^{-12} 	imes 10^5} = \frac{10^{-10}}{10^{-7}} = 10^{-3} \, T$.Using the right-hand rule,since $\vec{v}$ is in $\hat{i}$ and $\vec{F}$ is in $\hat{j}$,the magnetic field $\vec{B}$ must be in the $\hat{k}$ direction (since $\hat{i} 	imes \hat{k} = -\hat{j}$,we check the direction: $\vec{F} = q(\vec{v} 	imes \vec{B}) \implies \hat{j} = \hat{i} 	imes \hat{B} \implies \hat{B} = \hat{k}$).

If a particle of charge $10^{-12} \, C$ moving along the $\hat{x}$ direction with a velocity $10^5 \, m/s$ experiences a force of $10^{-10} \, N$ in the $\hat{y}$ direction due to a magnetic field,then the minimum magnetic field is:

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