A 90-mm length of wire moves with an upward velocity of 35 m s⁻¹ between the poles of a magnet. The magnetic field is 80 mT directed to the right. If the resistance in the wire is 5.00 mΩ, what are the magnitude and direction of the induced current? (Answer: 50.4 A)  



Given:

Length of the wire = l = 90 mm = 0.09 m

Upward velocity of the wire = v = 35  m s⁻¹

Magnetic field = B = 80 mT = 80 ｘ 10⁻³ T

Resistance in the wire = R = 5.00 mΩ = 5 ｘ 10⁻³ Ω


To Find:

The magnitude of the induced current = I = ?

The direction of the induced current =  ?

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Solution:

According to Ohm's law

V = R I

or 

I = `\frac {V}{R}` = `\frac {Ɛ}{R}` ----------- (1)

where Ɛ value can be found from Motional e.m.f. formula

Ɛ = B l v --------------(2)

Putting equation (2) in equation (1) we get


I = `\frac {B l v }{R}`


by putting the corresponding values

I = `\frac {80 ｘ 10⁻³ T ｘ 0.09 m ｘ 35 m s⁻¹ }{5 ｘ 10⁻³ Ω}`

I = `\frac {252 ｘ 10⁻³ T m² s⁻¹ }{5 ｘ 10⁻³ Ω}`

I = 50.4 A

Using Fleming's Right Hand Rule, the Direction of Induced Current will be directed into the plane of the paper.


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