A turntable has a moment of inertia of 3.00 x 10-2 kgm2 and spins freely on a frictionless bearing at 25.0 rev/min. A 0.300 kg b
1 answer:
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Answer:
3A is the larger of the two currents.
Explanation:
Let the currents in the two wires be I₁ and I₂
given:
Magnitude of the electric field, B = 4.0μT = 4.0×10⁻⁶T
Distance, R = 10cm = 0.1m
Ratio of the current = I₁ : I₂ = 3 : 1
Now, the magnitude of a magnetic field at a distance 'R' due to the current 'I' is given as
Where is the magnitude constant = 4π×10⁻⁷ H/m
Thus, the magnitude of a magnetic field due to I₁ will be
given,
B = B₁ - B₂ (since both the currents are in the same direction and parallel)
substituting the values of B, B₁ and B₂
we get
4.0×10⁻⁶T = -
or
4.0×10⁻⁶T =
also
⇒
substituting the values in the above equation we get
4.0×10⁻⁶T =
⇒
also
⇒
⇒
Hence, the larger of the two currents is 3A
Answer:
d) 1.2 mT
Explanation:
Here we want to find the magnitude of the magnetic field at a distance of 2.5 mm from the axis of the coaxial cable.
First of all, we observe that:
- The internal cylindrical conductor of radius 2 mm can be treated as a conductive wire placed at the axis of the cable, since here we are analyzing the field outside the radius of the conductor. The current flowing in this conductor is
I = 15 A
- The external conductor, of radius between 3 mm and 3.5 mm, does not contribute to the field at r = 2.5 mm, since 2.5 mm is situated before the inner shell of the conductor (at 3 mm).
Therefore, the net magnetic field is just given by the internal conductor. The magnetic field produced by a wire is given by
where
is the vacuum permeability
I = 15 A is the current in the conductor
r = 2.5 mm = 0.0025 m is the distance from the axis at which we want to calculate the field
Substituting, we find: