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2 answers:
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Answer:
the magnitude of the torque on the permanent magnet = 7.34×10⁻³ Nm
the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils = -1.0485 ×10⁻² J
Explanation:
The torque is given by :
where ;
m = 0.160 A.m²
B = 0.0800 T
θ = 35°
So the magnitude of the torque N = mBsinθ
N = (0.160)(0.0800)(sin 35°)
N = 0.007341
N = 7.34×10⁻³ Nm
Hence, the magnitude of the torque on the permanent magnet = 7.34×10⁻³ Nm
b) The potential energy
U = -mBcosθ
U = (- 0.160)(0.0800)(cos 45)
U = -0.010485
U = -1.0485 ×10⁻² J
Thus, the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils = -1.0485 ×10⁻² J
Answer:
Explanation:
charge, q = 10 C
time, t = 2 micro second
Current, i = q / t
i = 10 / (2 x 10^-6) = 5 x 10^6 A
(a)
distance, d = 1 m
the formula for the magnetic field is given by
B = 1 Tesla
Now the distance is d' = 1 km = 1000 m
B' = 0.001 Tesla
(b) The magnetic field of earth is Bo = 3 x 10^-5 tesla
B / Bo = 3.3 x 10^4
B'/Bo = 33.3
