Q. 74.1( 22 Votes )

# A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m s^{–1}, at right angles to the horizontal component of the earth’s magnetic field, 0.30 × 10^{–4} Wb m^{–2}.

(a) What is the instantaneous value of the emf induced in the wire?

(b) What is the direction of the emf?

(c) Which end of the wire is at the higher electrical potential?

Answer :

Given:

Length of the wire, l = 10 m

Speed of wire, v = 5.0 ms ^{- 1}

Magnetic field B = 0.30 × 10^{–4} Wb m^{–2}

The diagram is:

(a) The e.m.f induced in the wire can be calculated as follows:

e = Blv ...(1)

where B is the magnetic field

l is the length of the loop and,

v is the velocity of the rectangular loop

substituting the values in above equation, we get,

e = 0.30 × 10^{–4} Wb-m ^{- 2} × 10 m × 5ms^{-1}

e = 1.5 × 10^{–3} V

(b) The direction of the induced emf will be from west to east in accordance with the Fleming’s right - hand rule.

According to Fleming’s right - hand rule: When the right hand is held with the thumb, first finger and second finger mutually perpendicular to each other, as shown below.

Then, the thumb points in the direction in which the conductor moves; the first finger gives the direction of the magnetic field (N ⇒ S); and the second finger gives the direction of the induced current.

(c) The east end of the straight wire will be having high electrical potential.

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