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In the given figure, I was asked to calculate induced emf in the loop and individual wires of the triangle.

I have a doubt that when I am calculating the emf of the whole loop, between which two points I have to calculate, or I have to do something else.


motional emf = $\mathcal E=\displaystyle\oint(\vec v\times\vec B).\vec{dl}$

for arm AB induced emf will be zero because $(\vec v\times\vec B) $is $\perp$ to $\vec dl$

$\mathcal E_{ab}=0 $ (insert a battery of zero voltage in arm ab)

in arm BC,

$\mathcal E_{bc}=vBl_{bc}$ (insert a battery of this magnitude with it's positive terminal at B)

in $\triangle $ABC

let $\measuredangle CAB=\theta $ then velocity of arm AC can be resolved in two directions i.e,

$vcos\theta-$ along the arm AC

and ,

$vsin\theta-$ $\perp$ to the arm AC

the latter component of velocity will cause emf induced in it.

$\mathcal{E_{ac}}=vBl_{ac}sin\theta =vBl_{bc}=\mathcal{E_{bc}}$

(insert a battery of this magnitude with positive terminal at A and negative terminal at C)

now, apply KVL in loop ABC


so, in loop there will be no net emf induced.


The above-stated problem is based on motional EMF in which $$\mathcal{E}=\bigl(\overrightarrow{v}\times\overrightarrow{B}\bigr)\cdot\Delta\overrightarrow{l}$$ where $\Delta\overrightarrow{l}$ represents the projection of length along the vector perpendicular to $\overrightarrow{v}$. In this particular example, we have only two portions of conductor in which EMF will be induced, namely AC and BC(Notice that I'm putting higher potential points first).

There won't be any induced EMF in AB because $\bigl(\overrightarrow{v}\times\overrightarrow{B}\bigr)\cdot\Delta\overrightarrow{l}$ vanishes. Considering all this, we are left with: $$\mathcal{E}_{AC}=vBl_{BC}=-\mathcal{E}_{BC}$$ Therefore $\mathcal{E}_{total}=\mathcal{E}_{AC}+\mathcal{E}_{BC}=0$ .

This also complies with Faraday's Law which states that $$\frac{d\phi}{dt} = \oint{\overrightarrow{E}\cdot\overrightarrow{dl}}=\mathcal{E}_{total}$$ This is because the magnetic flux passing through the loop is constant, i.e $\dfrac{d\phi}{dt}=0$ , therefore no EMF is induced.

  • $\begingroup$ Current should flow in the direction CB. So, C should be at higher potential. $\endgroup$ – Yash Mittal Aug 19 '18 at 15:10
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    $\begingroup$ No, B will be at higher potential. $\endgroup$ – Utkarsh Verma Aug 19 '18 at 23:05
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    $\begingroup$ yes ,B will be at higher potential ,because in arm CB the magnetic force $q(\vec v \times \vec B)$ acts in downward direction (from C to B) which leads to the drift of positive charges from C to B . and to counter this magnetic force an electric field is devloped from B to C which in turn cause potential difference to be devloped between both ends (C,B) of arm CB $\endgroup$ – Faraday Pathak Aug 21 '18 at 5:32

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