Practice equation to find the potential difference of a circular loop moving in magnetic field, and motional EMF in conducting rod. Learn dependence of magnitude of induced EMF on angle between velocity and length of rod.
The emf generated due to the motion of rod is known as motional EMF. It is given by,
\(|\mathcal{E}| = B\,\ell v\)
where,
\(B\) is magnetic field
\(\ell\) is length of rod
\(v\) is velocity of rod and
\(\vec B, \vec \ell, \vec v\) are mutually perpendicular
The angle between velocity and length of rod is not 90º.
Thus, the component of velocity is so chosen that it should be perpendicular to the length of rod.
\(|\mathcal{E}| = v_{\bot} B\,\ell = B \,\ell\, v \,sin\, \theta\)
A \(\dfrac{B\ell v}{2}\)
B \(\dfrac{B\ell v}{3}\)
C \(\dfrac{B\ell v}{4}\)
D \(\dfrac{B\ell v}{5}\)
Since, there is no component of velocity which is perpendicular to the length.
Thus, there is no motional EMF.
Since, there is no component of velocity which is perpendicular to the length.
Thus, there is no motional EMF.
\(\mathcal{E} _{PQ} = B\,v \,dx\)
[\(\because\) length of PQ is \(dx\)]
\(v = x \,\omega\)
Thus, \(\mathcal{E} _{PQ} = B\,x \,\omega\, dx\)
The sum of induced emf of all small elements will be equal to the emf of rod.
Thus, \(\mathcal{E}_ {rod} = \displaystyle\int\limits^\ell_0 B \,x \,\omega \, dx\)
\(\mathcal{E} _{rod} = B\omega \left[\dfrac{x^2}{2}\right]^\ell_0\)
\(\mathcal{E} _{rod} = B \,\omega \dfrac{\ell^2}{2}\)
\(\mathcal{E} _{rod} = \dfrac{1}{2}\,B \,\omega\,\ell^2 \)
A \(\dfrac{1}{2}\, B\, \omega\, \ell^2\)
B \(\dfrac{1}{3}\, B\, \omega\, \ell^2\)
C \(\dfrac{3}{2}\, B\, \omega\, \ell^2\)
D \(2\, B\, \omega\, \ell^2\)
A \(\dfrac{B \omega}{2} (y^2 + 2\,xy)\)
B \(\dfrac{B \omega^2}{2} (x^2 + y^2)\)
C \(\dfrac{B \omega^2}{2} (x^2 + 2\,xy)\)
D \(B\, \omega^2\, x y\)
The potential difference between the points P and Q is given by,
\(V _{PQ} = B\, \ell \,v \) (as \(\vec B \,\bot\,\vec v\, \bot\,\vec \ell \))
= \(B × 2R× v\)
= \(2 \,B \, R \, v \)
A \(B\,R\,v\)
B \(2\,B\,R\,v\)
C \(\dfrac{B R v}{2}\)
D 0
\(\mathcal{E} _{PQ} = B\,v \,dx\)
[\(\because\) length of PQ is \(dx\)]
\(v = x \,\omega\)
Thus , \(\mathcal{E} _{PQ} = B\,x \,\omega\, dx\)
The sum of induced emf of all small elements will be equal to the emf of rod.
Thus, \(\mathcal{E}_ {rod} = \displaystyle\int\limits^\ell_0 B \,x \,\omega \, dx\)
\(\mathcal{E} _{rod} = B\omega \left[\dfrac{x^2}{2}\right]^\ell_0\)
\(\mathcal{E} _{rod} = B \,\omega \dfrac{\ell^2}{2}\)
\(\mathcal{E} _{rod} = \dfrac{1}{2}\,B \,\omega\,\ell^2 \)
A \(0.30\, V\)
B \(0.20\, V\)
C \(0.25 \,V\)
D \(0.16 \,V\)
\(\mathcal{E} _{PQ} = B\,v \,dx\)
[\(\because\) length of PQ is \(dx\)]
\(v = x \,\omega\)
Thus, \(\mathcal{E} _{PQ} = B\,x \,\omega\, dx\)
Thus, \(\mathcal{E}_ {rod} = \displaystyle\int\limits^\ell_0 B \,x \,\omega \, dx\)
\(\mathcal{E} _{rod} = B\omega \left[\dfrac{x^2}{2}\right]^\ell_0\)
\(\mathcal{E} _{rod} = B \,\omega \dfrac{\ell^2}{2}\)
\(\mathcal{E} _{rod} = \dfrac{1}{2}\,B \,\omega\,\ell^2 \)
A \(5 \,mV\)
B \(2.5 \,mV\)
C \(3 \,mV\)
D \(4 \,mV\)
A \(0.88\,V\)
B \(0.77\,V\)
C \(1.30\,V\)
D \(2.58\,V\)