An electron experiences a force vec{F} = (4.0 | vedclass

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Question

(A) The magnetic force is given by $\vec{F} = q(\vec{v} 	imes \vec{B})$. For an electron,$q = -e = -1.6 	imes 10^{-19}\,C$.Since the force is zero w

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$-0.075\,\hat{i} + 0.1\,\hat{j}$

Vedclass · Answer

(A) The magnetic force is given by $\vec{F} = q(\vec{v} 	imes \vec{B})$. For an electron,$q = -e = -1.6 	imes 10^{-19}\,C$.Since the force is zero when $\vec{v}_2 = (1.5\,\hat{i} - 2.0\,\hat{j}) 	imes 10^7\,m/s$,the magnetic field $\vec{B}$ must be parallel to $\vec{v}_2$. Thus,$\vec{B} = k(1.5\,\hat{i} - 2.0\,\hat{j})$ for some constant $k$.Using the first case: $\vec{F} = -e(\vec{v}_1 	imes \vec{B})$.$(4.0\,\hat{i} + 3.0\,\hat{j}) 	imes 10^{-13} = -1.6 	imes 10^{-19} 	imes (2.5 	imes 10^7\,\hat{k} 	imes k(1.5\,\hat{i} - 2.0\,\hat{j}))$.$(4.0\,\hat{i} + 3.0\,\hat{j}) 	imes 10^{-13} = -1.6 	imes 10^{-19} 	imes 2.5 	imes 10^7 	imes k 	imes (1.5(\hat{k} 	imes \hat{i}) - 2.0(\hat{k} 	imes \hat{j}))$.$(4.0\,\hat{i} + 3.0\,\hat{j}) 	imes 10^{-13} = -4.0 	imes 10^{-12} 	imes k 	imes (1.5\,\hat{j} + 2.0\,\hat{i})$.Comparing components: $4.0 	imes 10^{-13} = -4.0 	imes 10^{-12} 	imes k 	imes 2.0 \implies k = -0.05$.Thus,$\vec{B} = -0.05(1.5\,\hat{i} - 2.0\,\hat{j}) = -0.075\,\hat{i} + 0.1\,\hat{j}\,T$.

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