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Question

(A) The magnetic field at the origin due to the wire at $x=1 \ cm$ is $B_0 = \frac{\mu_0 i}{2 \pi (0.01)}$ in the $+z$ direction (using the right-hand

Vedclass · Accepted Answer

$\frac{-e U B_0}{2 \sqrt{2}}(\hat{i}-\hat{j})$

Vedclass · Answer

(A) The magnetic field at the origin due to the wire at $x=1 \ cm$ is $B_0 = \frac{\mu_0 i}{2 \pi (0.01)}$ in the $+z$ direction (using the right-hand rule,$\hat{k}$).The magnetic field at the origin due to the wire at $x=-2 \ cm$ is $B' = \frac{\mu_0 i}{2 \pi (0.02)}$ in the $-z$ direction $(-\hat{k})$.The net magnetic field at the origin is $\vec{B}_{net} = B_0 \hat{k} - \frac{B_0}{2} \hat{k} = \frac{B_0}{2} \hat{k}$.The velocity of the electron is $\vec{v} = U \cos(45^{\circ}) \hat{i} + U \sin(45^{\circ}) \hat{j} = \frac{U}{\sqrt{2}} \hat{i} + \frac{U}{\sqrt{2}} \hat{j}$.The force on the electron is $\vec{F} = q(\vec{v} 	imes \vec{B}) = (-e) \left( \frac{U}{\sqrt{2}} \hat{i} + \frac{U}{\sqrt{2}} \hat{j} ight) 	imes \left( \frac{B_0}{2} \hat{k} ight)$.Calculating the cross product: $\hat{i} 	imes \hat{k} = -\hat{j}$ and $\hat{j} 	imes \hat{k} = \hat{i}$.$\vec{F} = -e \left( \frac{U B_0}{2 \sqrt{2}} (-\hat{j}) + \frac{U B_0}{2 \sqrt{2}} (\hat{i}) ight) = \frac{-e U B_0}{2 \sqrt{2}} (\hat{i} - \hat{j})$.

$(i)$	$(ii)$	$(iii)$
$(A). \frac{\mu_0 i}{r} \otimes$	$(A). \frac{\mu_0 i}{4}(\frac{1}{r_1} - \frac{1}{r_2}) \otimes$	$(A). \frac{\mu_0 i}{4}(\frac{1}{r_1} - \frac{1}{r_2}) \otimes$
$(B). \frac{\mu_0 i}{2r} \odot$	$(B). \frac{\mu_0 i}{4}(\frac{1}{r_1} + \frac{1}{r_2}) \otimes$	$(B). \frac{\mu_0 i}{4}(\frac{1}{r_1} + \frac{1}{r_2}) \otimes$
$(C). \frac{\mu_0 i}{4r} \otimes$	$(C). \frac{\mu_0 i}{4}(\frac{1}{r_1} - \frac{1}{r_2}) \odot$	$(C). \frac{\mu_0 i}{4}(\frac{1}{r_1} - \frac{1}{r_2}) \odot$
$(D). \frac{\mu_0 i}{4r} \odot$	$(D). 0$	$(D). 0$

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