a bar is moving over two parallel rails with v=5m/s to the right perpendicular t

Question

a bar is moving over two parallel rails with v=5m/s to the right perpendicular to the magnetic field that is coming out of the page. the rails are connected through a resistor with R= 1 ohm. the rails and the bar themselves have no resistance. what is the magnitude and direction of the induced current in R when
a) the bar is at the left of R (moving towards R)
b) when the bar is at the right of R (moving away from R).
what is the magnetic force and it's direction on the bar in both cases?

ar to a magnetic field nd the bar A bar is moving over two parallel rails with v-5 m/s to the right perpendicular to a magnetic field that is coming out of the page. The rails are connected through a resistor with R=l ohm , The rails and themselves have no resistance. What is the magnitude and direction of the induced current in R when 1 the bar is at the left of R (moving towards R) b) a) the bar is at the right of R (moving away from it) What is the magnetie force, (magnitude and direction) on the bar in both cases? (By finding magnetic force you find the external force needed to keep the bar moving at constant v) 5% a, Asolenoid oflength 1=10 m and cross-section area A=5Wcmz has N:500 turns and carries a current

Explanation / Answer

a)

when the bar is at the left of R and moving towards R ::

E = induced EMF

V = speed of bar

B = magnetic field

induced current is given as

i = E/R = BLV/R = 3 x 0.5 x 5 / 1 = 7.5 A

direction of current = counterclockwise   since as the baar moves towards the resistance R , magnetic flux linked decreases. hence thebar experience forece in opposite direction that is left. then Using right hand rule , we get the direction of current as counterclockwise

magnetic force is given as

F = i BL = 7.5 x 3 x 0.5 = 11.25 N

direction : left

b)

when the bar is at the right of R and moving away R ::

E = induced EMF

V = speed of bar

B = magnetic field

induced current is given as

i = E/R = BLV/R = 3 x 0.5 x 5 / 1 = 7.5 A

direction of current = clockwise   since as the bar moves away from the resistance R , magnetic flux linked increases. hence the bar experience force in opposite direction that is left. then Using right hand rule , we get the direction of current as clockwise

magnetic force is given as

F = i BL = 7.5 x 3 x 0.5 = 11.25 N

direction : left

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