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# A vertical rod of length $l$ is moved with constant velocity $v$ towards the east. The vertical component of Earth’s magnetic field is $B$ and angle of dip is$\theta$. The induced E.M.F. in the rod is:(A)$Blv\cot \theta$(B)$Blv\sin \theta$(C)$Blv\tan \theta$(D)$Blv\cos \theta$

Last updated date: 23rd May 2024
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Hint: Whenever a body moves in the direction perpendicular to the magnetic field such that the vector obtained from the cross product of the velocity vector and the magnetic field vector is parallel to the length vector a potential difference is generated across both the ends of the rods which is known as induced E.M.F.

Formula used:
$E = (\overrightarrow v \times \overrightarrow B ) \bullet \overrightarrow l$
where $E$ is induced E.M.F., $\overrightarrow v$ is the velocity vector, $\overrightarrow B$ is the magnetic field vector, and $\overrightarrow l$ is the length vector.

From the question, we can say that the figure looks something like this,

where ${B_h}$ is the horizontal component of the magnetic field, $B$ is the vertical component of the magnetic field, and $v$ is the velocity with which the rod is travelling.
This ${B_h}$ is acting out of the plane towards the reader.
When we look from another angle the magnetic field looks like this,

Where $\theta$ is the angle of dip.
From the above figure, we can say that,
$\dfrac{{{B_h}}}{B} = \cot \theta$
And
$E = (\overrightarrow v \times \overrightarrow B ) \bullet \overrightarrow l$
where $E$ is induced E.M.F., $\overrightarrow v$ is the velocity vector, $\overrightarrow B$ is the magnetic field vector, and $\overrightarrow l$ is the length vector.
In the figure both the component of the magnetic field are perpendicular to the velocity vector of the rod but when we find the direction of the vector obtained through the cross product of velocity vector and vertical component of the magnetic field we see that it is perpendicular to the length vector of the rod so then their dot product will become zero (from the formula written above).
When we find the direction of the vector obtained through the cross product of velocity vector and the horizontal component of the magnetic field we see that it is parallel to the length vector of the rod so then their dot product will not be zero (from the formula written above).
Hence E.M.F. will only be induced due to the horizontal component of the magnetic field.
${B_h} = B\cot \theta$
Now using the formula,
$E = (\overrightarrow v \times \overrightarrow {{B_h}} ) \bullet \overrightarrow l$
$\Rightarrow E = v \times {B_h} \times l$
Hence
$E = v \times B\cot \theta \times l$
$\Rightarrow E = Bvl\cot \theta$

Hence the correct answer to the above question is (A) $Blv\cot \theta$

Note:
The direction of the resultant vector obtained from the cross product of two vectors was obtained using the right and palm rule according to which we have to keep our palm in direction of the first vector and curl the finger in the direction of the other vector and the third vector will give the direction of the resultant vector.