This question is incomplete, the complete question is;
Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to the magnetic field. The electric field strength 1.5 m from the center of the circle is 7 mV/m.
At what rate is the magnetic field changing?
Answer:
the magnetic field changing at the rate of 9.33 m T/s
Explanation:
Given the data in the question;
Electric field E = 7 mV/m
radius r = 1.5 m
Now, from Faraday law of induction;
∫E.dl = d∅/dt
E∫dl = A( dB/dt )
E( 2πr ) = πr² ( dB/dt )
( 0.007 ) = (r/2) ( dB/dt )
( 0.007 ) = 0.75 ( dB/dt )
dB/dt = 0.007 / 0.75
dB/dt = 0.00933 T/s
dB/dt = ( 0.00933 × 1000) m T/s
dB/dt = 9.33 m T/s
Therefore, the magnetic field changing at the rate of 9.33 m T/s
Explanation:
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Answer:
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Complete question:
What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field 0.425 T. (This is 60 rev/s.)
Answer:
The peak emf generated by the coil is 15.721 kV
Explanation:
Given;
Radius of coil, r = 0.250 m
Number of turns, N = 500-turn
time of revolution, t = 4.17 ms = 4.17 x 10⁻³ s
magnetic field strength, B = 0.425 T
Induced peak emf = NABω
where;
A is the area of the coil
A = πr²
ω is angular velocity
ω = π/2t = (π) /(2 x 4.17 x 10⁻³) = 376.738 rad/s = 60 rev/s
Induced peak emf = NABω
= 500 x (π x 0.25²) x 0.425 x 376.738
= 15721.16 V
= 15.721 kV
Therefore, the peak emf generated by the coil is 15.721 kV
Hello
1) First of all, since we know the radius of the wire (
), we can calculate its cross-sectional area
2) Then, we can calculate the current density J inside the wire. Since we know the current,
, and the area calculated at the previous step, we have
3) Finally, we can calculate the electric field E applied to the wire. Given the conductivity
of the aluminium, the electric field is given by
