Let's consider the cable's length finite, L. The total charge of the inner cylinder would be Q = λL. Applying Gauss' theorem ∫E*dA = λL/ε°, since E is uniform and perpendicular to dA: E∫dA = λL/ε° <=> E(2πrL) = λL/ε° <=> E = λ/2πε°r.
The units on both sides are also ok, so your formula is right.
If ρ is the resistance of the wire loop per unit length, and r is the radius of the loop, then the total resistance would be: R = ρ*(2πr). Applying Ohm's law, the induced current is: I = E/R. If you place the voltmeter, the measured value would be: V = I*ρ*l, where l is the arc length from the...
@dr2453, the voltage between battery terminals is constant, so current cannot flow across the capacitor. This is because the capacitor's plates are separated by a dielectric film or air, only a changing electric field (voltage) would make the current pass through it.
So, if a resistor it's...
(d) mentions a battery, so we are talking about dc current. In this case, the circuit is open at the capacitor. So, adding a resistor in series doesn't change anything, this eliminates (a). Neither does adding a resistor in parallel, because the voltage stays the same. Since you ruled out (c) by...