Question

A long metal bar of $30\text{cm}$ length is aligned along a north south line and moves eastward at a speed of $10{\text{ms}}^{-1}$. A uniform magnetic field of $4.0\text{T}$, points vertically downwards. If the south end of the bar has a potential of $0\text{V}$, the induced potential at the north end of the bar is

• A. $+12\text{V}$
• B. $-12\text{V}$
• C. $0\text{V}$
• D. cannot be determined since there is not closed circuit

Answer 1

The correct option is A $+12\text{V}$

The motional emf produced due to velocity $v$ of rod is given by
$\epsilon =vB{l}_{\perp }...\left(i\right)$
where $l_{\perp }=\text{length of rod perpendicular to its velocity}$
$\Rightarrow$since length of rod is making angle ${90}^{\circ }$ with east (Direction of velocity of rod), hence
$\therefore l_{\perp }=l=30\text{cm}=0.30\text{m}...\left(ii\right)$
Substituting values:
$\epsilon =vBl$
$\therefore \epsilon =10\times 4\times 0.30=12\text{V}$
$\Rightarrow$The sign of Motional emf $\left(\epsilon \right)$ is given by:
$\overrightarrow{v}\times \overrightarrow{B}$, so using right hand rule for cross product,the direction of thumb will point towards $+ve$ terminal.
$\Rightarrow$direction of $\overrightarrow{v}$ is $\hat{i}$, direction of $\overrightarrow{B}$ is $-\hat{k}$
$\therefore \hat{i}\times -\hat{k}=\hat{j}$ i.e along $+vey-$ direction (NORTH) for $(+ve)$ terminal of motional emf.
$\epsilon =V_N-V_S$
$12\text{V}=V_N-0$
$\therefore V_N=+12\text{V}$