When a thin 20.0cm iron rod moves with a constant velocity of 4.00m/s perpendicu

Question

When a thin 20.0cm iron rod moves with a constant velocity of 4.00m/s perpendicular to the rod in the direction shown in figure , the induced emf across its ends is measrued to be .300V

Part a

What is the magnitude of the magnetic field

B=?

Part B which point is at a higher potential A or B

Part C

If the bar is rotated clockwise by 90.0 degrees in the plane of the paper, but keeps the same veloicty, what is the potential difference induced across its ends.?

Vab=?V

When a thin 20.0cm iron rod moves with a constant velocity of 4.00m/s perpendicular to the rod in the direction shown in figure , the induced emf across its ends is measured to be .300V What is the magnitude of the magnetic field B=? If the bar is rotated clockwise by 90.0 degrees in the plane of the paper, but keeps the same velocity, what is the potential difference induced across its ends.?

Explanation / Answer

length of the rod is L = 20.0 cm = 20.0 x 10^-2 m

the speed of the rod is v = 4.00 m/s

let m be the mass of the rod and I be the current flowing through it

the weight of the rod is

w = m x g

where g = 9.8 m/s^2

the magnetic force on the rod is

F = B x I x L x sinA

where A = 90o

here,w = F

or B x I x L = m x g

or B = (m x g/I x L)

where B is the magnetic field

let E be the electric field

we know that

v = E/B

or E = v x B

the potential difference is

Vab = E x L

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