There is a uniform magnetic field B normal to | vedclass

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

Question

(C) Let the magnetic field be $\vec{B} = B \hat{k}$.The motional electromotive force $(EMF)$ induced in a conductor moving in a magnetic field is give

Vedclass · Accepted Answer

$v_x l_2 - v_y l_1$

Vedclass · Answer

(C) Let the magnetic field be $\vec{B} = B \hat{k}$.The motional electromotive force $(EMF)$ induced in a conductor moving in a magnetic field is given by $\int (\vec{v} 	imes \vec{B}) \cdot d\vec{l}$.For segment $AB$,the length vector is $\vec{l}_{AB} = l_1 \hat{i}$.The potential difference $V_A - V_B = \int_A^B (\vec{v} 	imes \vec{B}) \cdot d\vec{l} = (v_x \hat{i} + v_y \hat{j}) 	imes (B \hat{k}) \cdot (l_1 \hat{i}) = (-v_x B \hat{j} + v_y B \hat{i}) \cdot (l_1 \hat{i}) = v_y B l_1$.So,$V_A - V_B = B v_y l_1$.For segment $BC$,the length vector is $\vec{l}_{BC} = l_2 \hat{j}$.The potential difference $V_B - V_C = \int_B^C (\vec{v} 	imes \vec{B}) \cdot d\vec{l} = (v_x \hat{i} + v_y \hat{j}) 	imes (B \hat{k}) \cdot (l_2 \hat{j}) = (-v_x B \hat{j} + v_y B \hat{i}) \cdot (l_2 \hat{j}) = -v_x B l_2$.So,$V_B - V_C = -B v_x l_2$.Adding the two potential differences:$(V_A - V_B) + (V_B - V_C) = B v_y l_1 - B v_x l_2$.$V_A - V_C = B(v_y l_1 - v_x l_2)$.The magnitude of the potential difference $|V_A - V_C| = B |v_x l_2 - v_y l_1|$.Thus,the potential difference is proportional to $v_x l_2 - v_y l_1$.

Similar Questions

$A$ conductor $ABOCD$ moves along its bisector with a velocity of $1\, m/s$ through a perpendicular magnetic field of $1\, wb/m^2$,as shown in the figure. If all the four segments $(OB, BC, OC, CD)$ are of $1\, m$ length each,then the induced emf between points $A$ and $D$ is......$volt$.

$A$ conducting rod $AC$ of length $4l$ is rotated about a point $O$ in a uniform magnetic field $\vec B$ directed into the paper. $AO = l$ and $OC = 3l$. Then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module