Question

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.}

Hint:

Use Faraday's law to calculate the induced emf.

Answer 1

Correct Answer: 10

A conductor frame ABCDE, with each arm measuring 10 cm, moves rightward at a speed \(v=10~\text{cm s}^{-1}=0.1~\text{m s}^{-1}\) within a uniform magnetic field \(B=\dfrac{1}{\sqrt{2}}~\text{T}\) oriented perpendicularly to its plane. The objective is to determine the induced emf between points A and E.

Concept Used:

For a rigid conductor moving at a uniform velocity \(\vec v\) in a uniform field \(\vec B\) perpendicular to its plane, the motional electric field is defined as \(\vec E_m=\vec v\times\vec B\) (uniform). Consequently, the emf between two points is solely dependent on their displacement vector \(\vec{AE}\):

\[ \varepsilon_{AE}=\int_A^E (\vec v\times \vec B)\cdot d\vec \ell=(\vec v\times \vec B)\cdot \vec{AE}=Bv\,\Delta y, \]

where \(\Delta y\) represents the vertical separation between A and E. This is because \(\vec v\) is horizontal and \(\vec B\) is perpendicular to the plane, resulting in \(\vec v\times\vec B\) being vertical.

Step-by-Step Solution:

Step 1: Determine the vertical separation between A and E based on the provided geometry.

Arms AB and DE are horizontal and thus contribute no vertical displacement. The two angled arms, BC and CD, each form a \(45^\circ\) angle. Their respective vertical drops are calculated as \(10\cos45^\circ=\dfrac{10}{\sqrt2}\) cm. Therefore, the total vertical separation is:

\[ \Delta y = \frac{10}{\sqrt2}+\frac{10}{\sqrt2}= \frac{20}{\sqrt2}=10\sqrt2~\text{cm}=0.1\sqrt2~\text{m}. \]

Step 2: Apply the formula \(\varepsilon=Bv\Delta y\).

Substituting the given values:

\[ \varepsilon = \left(\frac{1}{\sqrt2}\right)(0.1)\left(0.1\sqrt2\right)=0.01~\text{V}. \]

Final Computation & Result:

The calculated induced emf between points A and E is:

\[ \boxed{\varepsilon=0.01~\text{V}=10~\text{mV}}. \]