Question

At a certain place, Earth's magnetic field has magnitude B =0.703 gauss and is inclined downward at an angle of 75.4° to the horizontal. A flat horizontal circular coil of wire with a radius of 10.0 cm has 1300 turns and a total resistance of 99.4 Ω. It is connected to a meter with 202 Ω resistance. The coil is flipped through a half-revolution about a diameter, so that it is again horizontal. How much charge flows in coulombs through the meter during the flip?

Answer 1

Answer:

The charge flows in coulombs is

$dq=1.843x10^{-5}C$

Explanation:

The current magnitude of current is given by the resistance and the induced Emf as:

$I=N*\frac{dF}{Rdt}$

$\frac{dq}{dt}=\frac{dF}{Rdt}=dq=N*\frac{dF}{R}$

$dq=\frac{N*\beta*A*(Cos(\alpha_f)-Cos(\alpha_i}{R}$

$N=1300$, $\beta=0.703$, $A=\pi*r^2=\pi*0.10^2=0.01\pi m^2$, $R=99.4+202=301.4$Ω

$\alpha_f=14.6$,$\alpha_i=165.4$

Replacing :

$dq=\frac{1300*0.703x10^{-4}*0.01\pi*(0.9667-(-0.9667))}{202+99.4}$

$dq=1.843x10^{-5}C$