The figure shows a stationary conductor whose shape is similar to the letter e.

Question

The figure shows a stationary conductor whose shape is similar to the letter e. The radius of its circular portion is a = 50.0 cm. It is placed in a constant magnetic field of 0.500 T directed out of the page. A straight conducting rod, 50.0 cm long, is pivoted about point O and rotates with a constant angular speed of 2.00 rad/s. (a) Determine the induced emf in the loop POQ. Note the area of the loop is a 2/2. (b) If all the conducting material has resistance per length of 5.00 /m, what is the induced current in the loop POQ at the instant 0.250 s after point P passes point Q?

Explanation / Answer

(a) Write the expression: = -d/dt
Again -
= B A
d/dt = B dA/dt (B is constant)
And -
A = (1/2)r²
dA/dt = (1/2)r²(d/dt) (r is constant, as is d/dt)
dA/dt = (1/2)r²()

Therefore -
= -(1/2)Br²()
= -(1/2)(0.500 T)(0.50 m)²(2.00 rad/s)
= -0.125 V

(b) Here, there 's no information here about the conducting rod at t = 0 (initial position).
If it starts from OQ,
= t
= (2.00 rad/s)(0.250 s)
= 0.50 rad

Now -
L = 2r + r
L = r(2 + )
L = (0.50 m)(2 + 0.50 rad)
L = 1.25 m

R = (1.25 m)(5.0 /m)
R = 6.25

Therefore -
V = IR
0.125 V = I(6.25 )
=> I = 0.0200 A = 20.0 mA

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