An electrical appliance has a resistance of 40 Ω. When this electrical appliance is connected to a 220 V supply line, the curren
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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 with Explanation:
We are given that
Force acts on a pebble= N
Position vector= m
a.We have to find the resulting torque on the pebble about origin.
Torque=
Substitute the values then we get
Torque=N-m
By using
b.
Torque about point (2,0,-3)
N-m