A rectangular coil with resistance 200ohms hs 50 turns each each of length l=10c

Question

A rectangular coil with resistance 200ohms hs 50 turns each each of length l=10cm and width w=5cm. the coil moves into a uniform magnatic feild B=1.2T with constant velocity V=10m/s.

In all cases calculate the magnitude And the direction

a)Calculate the induced emf when the right edge is about to penetrate the feild.

b) Calculate the induced emf when the left edge of the coil is about to penetrate the feild.

c)Calculate the induced emf when the left edge of the ccoil moves from beig just totally inside the feild (on the left side) until it is just about to emerge from the feild on the right.

d)Calculate the force on each of the edges of the coil.

i)Whule the coil is penetrating the feild and ii) while it is totally inside the feild.

e) now the coil is lying motionless and flat within the feild and it is rotated about an axes through the right side of the coil, such that the left side rises, leaving the coil prependicular to the feild. this is done in 0.1 seconds. what is the average emf induced in the coil and what is the average induced current.

A rectangular coil with resistance 200ohms hs 50 turns each each of length l=10cm and width w=5cm. the coil moves into a uniform magnetic feild B=1.2T with constant velocity V=10m/s. In all cases calculate the magnitude And the direction Calculate the induced emf when the right edge is about to penetrate the feild. Calculate the induced emf when the left edge of the coil is about to penetrate the feild. Calculate the induced emf when the left edge of the ccoil moves from beig just totally inside the feild (on the left side) until it is just about to emerge from the feild on the right. Calculate the force on each of the edges of the coil. Whule the coil is penetrating the feild and ii) while it is totally inside the feild. now the coil is lying motionless and flat within the feild and it is rotated about an axes through the right side of the coil, such that the left side rises, leaving the coil perpendicular to the feild. this is done in 0.1 seconds. what is the average emf induced in the coil and what is the average induced current.

Explanation / Answer

a) E = nBlv = 1.2*50*0.05*10=30V current will flow counterclokwise.

b) E = nBlv=30 V the current will flow counterclokwise

c) E = nBlv = 30 V the current will flow clokwise

d)i) while penetrating the field

force on right edge = nILB = 50*(30/50)*0.05*1.2= 1.8 N towards left

force on upper edge = 0

force on lower edge = 0

force on left edge = 0

ii) while inside the field

force on right edge = nILB = 50*(30/50)*0.05*1.2= 1.8 N towards left

force on upper edge = nILB = 50*(30/50)*0.10*1.2= 3.6 N downward

force on lower edge = nILB = 50*(30/50)*0.10*1.2= 3.6 N upward

force on left edge = nILB = 50*(30/50)*0.05*1.2= 1.8 N towards right

e)E = -dB/dt = -(0-1.2*50*0.10*0.05)/0.1= 3V

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