A conducting semi-circular loop of radius \( R = 0.1 \, \text{m} \), with its diameter centered at the origin, rotates in the \( x \)-\( y \) plane about the origin with a constant angular velocity, \( \omega = 20 \, \text{rad/s} \), as shown. A magnetic field of magnitude \( B = 2 \, \text{T} \) and normal to the \( x \)-\( y \) plane exists in the region \( x \geq 0 \) as shown. If the loop has a resistance of \( 2 \, \Omega \), and negligible inductance, the peak-to-peak current (in milliamperes) in the loop is _________ (round off to one decimal place)
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The current induced in a rotating loop in a magnetic field is given by \( I = \frac{B A \omega}{R} \), where \( B \) is the magnetic field, \( A \) is the area of the loop, \( \omega \) is the angular velocity, and \( R \) is the resistance of the loop.
The induced EMF in the rotating loop is given by the formula for the motional EMF:
\[
\mathcal{E} = B A \omega
\]
where:
- \( B = 2 \, \text{T} \) (magnetic field),
- \( A = \frac{1}{2} \pi R^2 \) (area of the semi-circular loop),
- \( \omega = 20 \, \text{rad/s} \) (angular velocity).
First, calculate the area \( A \):
\[
A = \frac{1}{2} \pi (0.1)^2 = 0.005 \, \text{m}^2.
\]
Now, calculate the induced EMF:
\[
\mathcal{E} = (2) (0.005) (20) = 0.2 \, \text{V}.
\]
The current is given by Ohm's law:
\[
I = \frac{\mathcal{E}}{R} = \frac{0.2}{2} = 0.1 \, \text{A} = 100 \, \text{mA}.
\]
The peak-to-peak current is twice the peak current, so:
\[
I_{\text{peak-to-peak}} = 2 \times 100 = 200 \, \text{mA}.
\]
Thus, the peak-to-peak current is \( 200.0 \, \text{mA} \).
