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richard7893
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For Torque of current carrying wire in magnetic field why does the magnetic field of the current carrying wire itself not contribute to torque?
Torque of a current carrying wire in magnetic field
AI Thread Summary
The torque on a current-carrying wire in a magnetic field is not influenced by the wire's own magnetic field because that field does not exert a force on itself. The magnetic field generated by the wire circulates around it and does not interact directly with the wire to produce torque. Consequently, the wire does not experience a force from its own magnetic field. This principle is fundamental in understanding electromagnetic interactions. Overall, the current-carrying wire only responds to external magnetic fields for torque generation.
Neglect gravity other than that of water; neglect friction.
F1=ρgsh=1000*10*1*(1+4)=50000 N;
F1=ρgsh=1000*10*0.1*4=4000 N;
F3=50000-4000=46000 N?
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system
$$M(t) = M_{C} + m(t)$$
$$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$
$$P_i = Mv + u \, dm$$
$$P_f = (M + dm)(v + dv)$$
$$\Delta P = M \, dv + (v - u) \, dm$$
$$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$
$$F = u \frac{dm}{dt} = \rho A u^2$$
from conservation of momentum , the cannon recoils with the same force which it applies.
$$\quad \frac{dm}{dt}...
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