Heart-lung machines and artificial kidney mac | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$JhB$

Vedclass · Answer

(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.

Similar Questions

Two very long straight parallel conductors of negligible cross-section carrying equal electric current of $1 \text{ mA}$ are placed one metre apart in a vacuum. The force produced per metre length is $\qquad \text{ N}$. (Given: $\mu_0 = 4\pi \times 10^{-7} \text{ T m A}^{-1}$)

Two long and parallel straight wires $A$ and $B$ carrying currents of $8.0 \, A$ and $5.0 \, A$ in the same direction are separated by a distance of $4.0 \, cm$. Estimate the force on a $10 \, cm$ section of wire $A$.

$A$ wire carrying a current $I$ along the positive $x$-axis has length $L$. It is kept in a magnetic field $\overrightarrow{B} = (2\hat{i} + 3\hat{j} - 4\hat{k}) \text{ T}$. The magnitude of the magnetic force acting on the wire is $..........IL$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module