Answer: 2.6*10^-8 Ωm
Explanation:
the length of the wire, l = 12.2 m
the width of the wire, w = 1.8 mm
the thickness of the wire, T = 0.11 mm
the potential difference of the battery, v = 12 V
the current in the battery, I = 7.5 A
Recall, Ohm’s law states, V = IR.
Hence, R = V/I
R = 12/7.5 = 1.6 Ω
Resistivity of a material is given by
ρ = RA/l
ρ = [1.6 * 1.8*10^-3 * 0.11*10^-3] / 12.2
ρ = (3.168*10^-7) / 12.2
ρ = 2.596*10^-8
Thus, the resistivity of the wire is 2.60*10^-8 Ωm
a) (0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^) N.m
b) ΔU = -0.000747871 J
c) w = 47.97 rad/s
The relevant data includes the area of the circular ring as 4.45 cm², the current of 13.5 Amps, and the magnetic field strength as (1.05×10−2T)(12i^ + 3j^ - 4k^). The initial magnetic moment orientation is expressed as μi = μ(−0.8i^ + 0.6j^). The moment of inertia for the ring is given as 6.50×10−7 kg⋅m². To find the initial magnetic moment (μ), we apply μ = N*I*A, using N=1 for a single coil. Substituting the values yields μ = 0.0060075 A-m². The torque on the ring is determined using the cross product of the magnetic moment and the magnetic field. Calculating this gives the initial torque as (0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^). The magnetic potential energy is then calculated using the dot product of the magnetic moment with the magnetic field. This leads to initial energy stored in the ring as Ui = 0.000495556 J after computation. Following a 90-degree rotation, the final potential energy is found to be Uf = -0.000252315 J. The decrease in potential energy is determined to be ΔU = -0.000747871 J. Thus, the potential energy stored in the ring decreased by ΔU = -0.000747871 J.
