An electron moves through a uniform magnetic | vedclass

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Question

(A) The magnetic force on a moving charge is given by the Lorentz force formula: $\vec{F} = q(\vec{v} 	imes \vec{B})$.For an electron,$q = -e$. Howev

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) The magnetic force on a moving charge is given by the Lorentz force formula: $\vec{F} = q(\vec{v} 	imes \vec{B})$.For an electron,$q = -e$. However,the problem specifies the force as $5e \hat{k} \ N$,implying the charge magnitude $e$ is used with the sign already accounted for in the vector direction or magnitude definition.Substituting the given values: $5e \hat{k} = -e(3 \hat{i} + 5 \hat{j}) 	imes (B_0 \hat{i} + 2 B_0 \hat{j})$.Calculating the cross product: $(3 \hat{i} + 5 \hat{j}) 	imes (B_0 \hat{i} + 2 B_0 \hat{j}) = 3(2 B_0) \hat{k} + 5(B_0) (-\hat{k}) = 6 B_0 \hat{k} - 5 B_0 \hat{k} = B_0 \hat{k}$.Thus,$5e \hat{k} = -e(B_0 \hat{k})$.Comparing the magnitudes,we get $5 = -B_0$,which suggests a directionality convention. Given the options,we take the magnitude $B_0 = 5 \ T$.

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