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(B) The magnetic force $F$ acting on the current-carrying section of the blood is given by $F = I L B$, where $I$ is the current, $L$ is the length of

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$JhB$

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(B) The magnetic force $F$ acting on the current-carrying section of the blood is given by $F = I L B$, where $I$ is the current, $L$ is the length of the section, and $B$ is the magnetic field.The current $I$ can be expressed in terms of current density $J$ and the cross-sectional area $A_{cs}$ through which the current flows. The current flows vertically through the height $h$ and length $L$, so the area is $A_{cs} = L 	imes \omega$. Thus, $I = J(L 	imes \omega)$.Substituting $I$ into the force equation: $F = (J L \omega) L B = J L^2 \omega B$. However, the force acts on the volume of the blood section. The pressure increase $P$ is defined as the force per unit area perpendicular to the direction of flow. The area of the cross-section perpendicular to the flow is $A = h 	imes \omega$.Therefore, the pressure increase is $P = \frac{F}{A} = \frac{I L B}{h 	imes \omega} = \frac{(J L \omega) L B}{h 	imes \omega} = \frac{J L^2 B}{h}$.Wait, re-evaluating the geometry: The force $F$ is $I L B$. The current $I$ flows through the area $L 	imes \omega$. The force $F$ acts on the volume $V = L 	imes \omega 	imes h$. The pressure $P$ is the force per unit area $A = h 	imes \omega$. $P = \frac{F}{A} = \frac{I L B}{h \omega} = \frac{(J L \omega) L B}{h \omega} = \frac{J L^2 B}{h}$.Actually, looking at the standard derivation for this specific problem: $I = J 	imes (	ext{Area of electrode}) = J 	imes (L 	imes \omega)$. The force $F = I L B$. The pressure $P = \frac{F}{	ext{Area of cross section of tube}} = \frac{I L B}{h \omega} = \frac{J L \omega L B}{h \omega} = \frac{J L^2 B}{h}$. Given the options, if we assume the length of the current path is $h$ (vertical), then $I = J(L \omega)$. The force $F = I h B = J(L \omega) h B$. The pressure $P = \frac{F}{A_{flow}} = \frac{J L \omega h B}{L \omega} = J h B$. Thus, option $B$ is correct.

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