Prob. 4 A 1.2m-long metal bar is pulled to the right at a steady rate of 2.5m/s perpendicular to a uniform, 1.50T magnetic field. The bar rides on a pair of conducting, parallel rails connected through a resistance R = 7.50 so the apparatus forms a complete electrical circuit. Ignore the resistance of the bar and the rails and all gravitational effects. [b]What is the magnitude of the induced current (in units of A)?

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Prob. 4 A 1.2m-long metal bar is pulled to the right at a steady rate of 2.5m/s perpendicular to a
uniform, 1.50T magnetic field. The bar rides on a pair of conducting, parallel rails connected
through a resistance R = 7.5N so the apparatus forms a complete electrical circuit. Ignore the
resistance of the bar and the rails and all gravitational effects.
[b]What is the magnitude of the induced current (in units of A)?
Transcribed Image Text:Prob. 4 A 1.2m-long metal bar is pulled to the right at a steady rate of 2.5m/s perpendicular to a uniform, 1.50T magnetic field. The bar rides on a pair of conducting, parallel rails connected through a resistance R = 7.5N so the apparatus forms a complete electrical circuit. Ignore the resistance of the bar and the rails and all gravitational effects. [b]What is the magnitude of the induced current (in units of A)?
Expert Solution
Step 1: Rate of change of magnetic flux through a loop induces emf

As the rod moves with uniform velocity, the magnetic flux through the loop formed by the rod and the conducting parallel rails increases. It keeps on changing as the rod moves forward.

emf=-dϕdtφ=magnetic flux=B×L×vt

where B is the magnetic field

L is the length of the rod

v is the velocity of the rod

Therefore,

emf=-d(BLvt)dt=BLvmagnitude =Blv

 

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