A conducting rod of length L lies in the XY-p | vedclass

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(A) The rod of length $L$ makes an angle $30^{\circ}$ with the $X$-axis. The projection of the rod along the $Y$-axis is $l = L \sin 30^{\circ} = \fra

Vedclass · Accepted Answer

$\frac{B_0 v_0 L}{64}$

Vedclass · Answer

(A) The rod of length $L$ makes an angle $30^{\circ}$ with the $X$-axis. The projection of the rod along the $Y$-axis is $l = L \sin 30^{\circ} = \frac{L}{2}$.Since the rod moves with velocity $v_0$ along the $X$-axis,the motional emf is induced due to the component of the rod perpendicular to the velocity vector. The effective length perpendicular to the velocity $v_0$ (which is along the $X$-axis) is the projection of the rod along the $Y$-axis.Consider a small element of the rod of length $dy$ at a distance $y$ from the origin along the $Y$-axis. The magnetic field at this position is $B = B_0 \left(\frac{y}{L}\right)^3$.The induced emf $dE$ across this small element is given by $dE = B v_0 dy$.Substituting the expression for $B$:$dE = B_0 \left(\frac{y}{L}\right)^3 v_0 dy = \frac{B_0 v_0}{L^3} y^3 dy$.To find the total emf $E$ induced in the rod,we integrate $dE$ from $y = 0$ to $y = L/2$:$E = \int_0^{L/2} \frac{B_0 v_0}{L^3} y^3 dy = \frac{B_0 v_0}{L^3} \left[ \frac{y^4}{4} \right]_0^{L/2}$.$E = \frac{B_0 v_0}{L^3} \left( \frac{(L/2)^4}{4} - 0 \right) = \frac{B_0 v_0}{L^3} \left( \frac{L^4}{16 \cdot 4} \right) = \frac{B_0 v_0 L}{64}$.

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