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Question: A rod of length \(10cm\) made up of conducting and non conducting material (shaded part in non –cond...

A rod of length 10cm10cm made up of conducting and non conducting material (shaded part in non –conducting). The rod is rotted with constant angular velocity 10rad/sec10rad/\sec about point O, in contact magnetic field of 22 tesla as shown in figure. The induced Emf between the point A and B of rod will be?

Explanation

Solution

In this type of question first we have to analyze what is given in question and figure then, using given information and formulae for finding Emf of non conducting material will find the equation and then integrate using values for initial and final as given.

 ![](https://www.vedantu.com/question-sets/1ccc2681-5208-49d4-b653-631152d64ab8133316989470002928.png)  

Complete step-by-step answer:
Given length of the rod 1=10cm1=10cm
Magnetic field B=2TB=2T
And regular speed ω=10rad/sec\omega =10rad/\sec
Considering an element length dr at a distance X from end O.
Emf experienced by the elementary length of the conductor.
dε=BVdxd\varepsilon =BVdx
Where,
v=ωrv=\omega r
Thus the above equation becomes
dε=Bωrdrd\varepsilon =B\omega rdr
Now, potential difference between the point A and B is
ε=0.070.01Bwdr\varepsilon =\int{_{0.07}^{0.01}Bwdr}
Simplifying it by solving integration,
We get,
=Bwr20.0720.1=Bw{{\dfrac{r}{{{2}_{0.07}}}}^{{{2}^{0.1}}}}
Solving,
2×10×(0.120.072)22\times 10\times \dfrac{\left( {{0.1}^{2}}-{{0.07}^{2}} \right)}{2}
Therefore the final answer is
0.051V\Rightarrow 0.051V

Note: In this type of problem remember the formulae and values for integration.
Note that it is used to find the length of the conductor.