Question

a wire 5 m long is supported horizontally at a height of 15m along east west direction.when it is about to hit the ground what will be the average emf induced in it. the wire falls freely in the field of the horizontal component of the earth's magnetic induction of 3.6 x 10-5t​

Answer 1

According to Fleming's right hand rule, if we stretch the first finger, central finger and thumb of our right hand in mutual direction such that first finger points along the direction of the field and thumb is along the direction of the motion of the conductor, then the central finger would give us the direction of the induced current. Here the wire is falling vertically, therefore the direction of motion of the conductor that is the thumb is in a downward direction.

The magnetic field produced that is the index finger is in the direction of the horizontal component of earth. the induced emf which is the middle finger is perpendicular to both.

Therefore, it is perpendicular to the plane containing the horizontal component and the direction of motion.

Answer 2

when wire is about to hit the ground  the average emf induced in it is   $3.1176 X 10^{-3}$ V

given

length of wire = 5m

height at which it is supported = 15m

direction = east west

wire falls in B(h)= 3.6 x $10^{-5}$ Tesla

to find

average emf induced in it.

solution

• emf induced in a wire moving perpendicularly to the magnetic field is given by following expression:

E=BLV

B= magnetic field perpendicular to which the wire falls

L= length of wire

V= speed with which wire falls

• in the question the wire is place at east west and it falls down with the ends of the wire pointing east and west direction respectively
• E will only be induced if wire is falling perpendicularly to the magnetic field
• in the question it is given that wire is falling in horizontal component of earth's magnetic field , so the field is parallel to earth surface and perpendicular to east west direction

• velocity of wire is given by

$v^{2} - u^{2} = 2as$

v is the velocity of wire

u is the initial velocity of wire = 0 m/s

a is the acceleration of wire = $10 m/s^{2}$

s is the displacement of wire= 15 m

$v^{2} =2 X 10 X 15$

$v^{2} =300$

v=17.32 m/s

• so putting the values and  calculating E we get

E= 3.6 x $10^{-5}$  X   5  X   17.32

E = 311.76 X $10^{-5}$ = $3.1176 X 10^{-3}$ V

when wire is about to hit the ground   the average emf induced in it is E = $3.1176 X 10^{-3}$ V

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