Question

A uniform metal rod is moving with a uniform velocity v parallel to a long straight wire carrying a current I. The rod is perpendicular to the wire with its ends at distances r1 and r2 (with r2>r1) from it. The emf induced in the rod is

Answer 1

Final Answer :
$\frac{uiv}{2\pi} ln( \frac{r(2)}{r(1)} )$
where u is permutivity in vacuum
i is current.


Understanding :
1)Motional EMF
E = B. (V x L)
where v is velocity
L is length of particle having magnetic field B.


Steps :
1) By motion of Metallic rod (conductor) in magnetic field, EMF is generated in some conditions.
2) Since, Magnetic field due to Long straight wire varies with x, therefore we will find first emf at distance x of element dx of rod.
Then, we will integrate it from r1 to r2.

For Calculation see pic

Answer 2

Answer:

Dear Student,

Here we are using concept of Fleming's right hand rule.

Fleming’s Right Hand Rule is used to find the direction of the induced emf of a conductor moving in a magnetic field. This rule is also called the Generator rule. The thumb, the first and the second fingers on the right hand are held so that they are at right angles to each other. If the first finger points in the direction of the magnetic field and the thumb in the direction of the motion of the conductor then the second finger will point in the direction of the induced current in the conducted

According to Fleming's right hand rule Potential at A will be higher than that at B

option (d ) is correct

Regards