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

The charge flows in coulombs is

`dq=1.843x10^(-5)C`

Step-by-step explanation:

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

`I=N*(dF)/(Rdt)`

`(dq)/(dt)=(dF)/(Rdt)=dq=N*(dF)/(R)`

`dq=(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=(1300*0.703x10^(-4)*0.01\pi*(0.9667-(-0.9667)))/(202+99.4)`

`dq=1.843x10^(-5)C`