Conductor wire ABCDE with each arm 10 cm in length is placed in magnetic field of $\frac{1}{\sqrt{2}}$ Tesla, perpendicular to its plane. When conductor is pulled towards right with constant velocity of $10 \mathrm{~cm} / \mathrm{s}$, induced emf between points A and E is _______ mV.} 
Conductor wire ABCDE with each arm 10 cm in length is placed in magnetic field of $\frac{1}{\sqrt{2}}$ Tesla, perpendicular to its plane. When conductor is pulled towards right with constant velocity of $10 \mathrm{~cm} / \mathrm{s}$, induced emf between points A and E is _______ mV.}
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Correct Answer: 10
Approach Solution - 1
A conductor frame ABCDE (each arm 10 cm) moves to the right with speed \(v=10~\text{cm s}^{-1}=0.1~\text{m s}^{-1}\) in a uniform magnetic field \(B=\dfrac{1}{\sqrt{2}}~\text{T}\) perpendicular to the plane. We need the induced emf between A and E.
Concept Used:
For a rigid conductor moving with uniform velocity \(\vec v\) in a uniform field \(\vec B\perp\) plane, the motional electric field is \(\vec E_m=\vec v\times\vec B\) (uniform). Hence the emf between two points depends only on their displacement \(\vec{AE}\):
\[ \varepsilon_{AE}=\int_A^E (\vec v\times \vec B)\cdot d\vec \ell=(\vec v\times \vec B)\cdot \vec{AE}=Bv\,\Delta y, \]
where \(\Delta y\) is the vertical separation between A and E (since \(\vec v\) is horizontal and \(\vec B\) is perpendicular to the plane, \(\vec v\times\vec B\) is vertical).
Step-by-Step Solution:
Step 1: Find the vertical separation between A and E from the geometry.
AB and DE are horizontal (no vertical contribution). The two slanted arms BC and CD are at \(45^\circ\) each, so their vertical drops are \(10\cos45^\circ=\dfrac{10}{\sqrt2}\) cm each. Thus
\[ \Delta y = \frac{10}{\sqrt2}+\frac{10}{\sqrt2}= \frac{20}{\sqrt2}=10\sqrt2~\text{cm}=0.1\sqrt2~\text{m}. \]
Step 2: Use \(\varepsilon=Bv\Delta y\).
\[ \varepsilon = \left(\frac{1}{\sqrt2}\right)(0.1)\left(0.1\sqrt2\right)=0.01~\text{V}. \]
Final Computation & Result:
Therefore, the induced emf between A and E is
\[ \boxed{\varepsilon=0.01~\text{V}=10~\text{mV}.} \]
Approach Solution -2
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