# Please solve 9-10 and 11 I-9- In the figure to the left below a wire o.docx

8 Feb 2023
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### Please solve 9-10 and 11 I-9- In the figure to the left below a wire o.docx

• 1. Please solve 9,10 and 11 I.9] In the figure to the left below a wire of length L is immersed in a uniform magnetic field B out of the paper. The wire is moving at a speed v toward the right along two parallel rails connected to a resistance R. The current flowing in the circuit is (neglecting self inductance) X) counterclockwise BBLu clockwise [C] B v2 counterclockwise [D] PL2u counterclockwise [E]BR-clockwise i(t) z axis out of the paper 2 (t) 10 The figure at the center above shows an electric field increasing in time E,(t) = E0+E where o, E1, and T are all positive. The electric field is uniform in space with cylindrical symmctr direction shown at a distance r from the center of symmetry Usc Ampcre-Maxwell's law to find the induced magnetic field Bo in tho 2?? .11] In the AC R-L-C circuit shown to the right in the figure above the applied Emf is E(t) = m coswt. Keeping m constant, as the frequency ? increases, the rms current through the resistor A] increases B] decreases [C does n X may increase or decrease depending on the value of the resonant frequency. not change Solution 9) A) B*L*v/R counter clockwise Induced emf = B*L*v induced current = induced emf/R = B*L*v/R the direction of induced current is always such as to oppose the change in magnetic flux that generated it. so, induced current direction must be clouner clockwise. 10) B) Bc = mue*epsilon*pi*r^2*E1/T because, displacement current, Id = A*epsilon*dE/dt
• 2. = pi*r^2*epsilon*E1/T using Ampere-Maxwell's equation integral Bc*dL = mue*Id Bc*2*pi*r = mue*pi*r^2*epsilon*E1/T Bc = mue*epsilon*r*E1/(2*T) Bc = mue*epsilon*A*dE/dt = mue*epsilon*pi*r^2*E1/T 11) D) may increase or decrease depending on the value of the resonant frequency. at resonant frequency the rms current through the circuit is maximum. if the frequency of the source is less than or greater than resonant frequency the rms current decreases.