Magnetic Field and Loop 2: Direction and Induced Current

AI Thread Summary
The magnetic field in Loop 2 is directed towards the negative y-axis. If the resistance Rx is increasing, the current in Loop 1 decreases, resulting in a stronger magnetic field. According to Lenz's Law, the induced current in Loop 2 will flow in a clockwise direction. The discussion seeks confirmation on the correctness of these interpretations. The responses emphasize the relationship between resistance changes, magnetic fields, and induced currents.
frozen7
Messages
163
Reaction score
0
b) In Figure B2, a variable resistor Rx is connected to a DC voltage source Vs.
i) What is the direction of magnetic field act inside loop 2?
ii) If the Rx is decreasing, what is the direction of the induced current acting on loop 2? Explain your answer.

bi) The direction of the magnetic field in Loops 2 should be acting towards negative y-axis.

bii) If the Rx is increasing,then the current flow in loop 1 will be decreasing, hence the magnetic field is getting stronger as well. The current in loop2 will flow in clockwise direction according to Lenz`s Law.

Is my answer correct?
 
Physics news on Phys.org
Can anyone check my answer please?
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
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}...
Back
Top