A horizontal straight wire 10 m long extending from east to west isfalling with a speed of 5.0 m s–1, at right angles to the horizontalcomponent 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?
Given: The length of the wire is 10 m , the falling speed of the wire is 5.0 m/s and the magnetic field strength of the earth 0.3 × 10 − 4 Wbm − 2 .
a)
The induced emf is given as,
e = B l v
Where, the length of the wire is l , the magnetic field strength is B and the falling speed of the wire is v .
By substitute the given values in the above formula, we get
e = 0.3 × 10 − 4 × 10 × 5 = 1.5 × 10 − 3 V
Thus, the instantaneous value of the emf induced in the wire is 1.5 × 10 − 3 V .
b)
The direction of emf can be determined by the Fleming’s right hand rule. The Fleming’s right hand rule gives the direction of the induced current in a conductor moving in a magnetic field.
Thus, the wire is falling, so the direction of induced emf will be from West to East.
c)
From Fleming’s right hand rule, the direction of induced emf is from West to East, so the current will flow from Eastern end to western end.
Thus, the eastern end of the wire will be at higher potential.
Given: The length of the wire is
a)
The induced emf is given as,
Where, the length of the wire is
By substitute the given values in the above formula, we get
Thus, the instantaneous value of the emf induced in the wire is
b)
The direction of emf can be determined by the Fleming’s right hand rule. The Fleming’s right hand rule gives the direction of the induced current in a conductor moving in a magnetic field.
Thus, the wire is falling, so the direction of induced emf will be from West to East.
c)
From Fleming’s right hand rule, the direction of induced emf is from West to East, so the current will flow from Eastern end to western end.
Thus, the eastern end of the wire will be at higher potential.
