Antoha1
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- Homework Statement
- A frame (a loop) with an area of 5 cm^2 and a resistance of 2 Ω is placed in a uniform magnetic field with an induction (B) of 0.1T. A galvanometer is connected to the frame. What charge (q) will flow through the galvanometer when the frame is rotated through an angle of ##\beta##=120 degrees? At the beginning of the observation, the plane of the frame is perpendicular to the lines of magnetic field.
- Relevant Equations
- $$\phi=BAcos\theta$$
$$\Delta\phi=\phi_{1}-\phi_{2}=BA_{1}-BA_{2}=B(A_{1}-A_{2})=B(Acos\alpha+Acos(\alpha+\beta))$$
$$\varepsilon=\frac{\Delta\phi}{\Delta t}=IR=\frac{\Delta q}{\Delta t}R$$
##\varepsilon## - EMF
from the written formulas, charge is:
$$\Delta q=\frac{BA}{R}(cos\alpha-cos(\alpha+\beta))$$ .
It is said that in the beginning, the Area was perpendicular to magnetic field lines so the
##\alpha=0##
##\phi_{1}=BAcos0=BA## ,
Refered to that:
$$\Delta q=\frac{BA}{R}(1-cos\beta)$$
but, thinking logically, cosine function becomes negative at 120 degrees so the subtraction of cosines becomes the sum. And I do not think that ##\Delta q## can be bigger than BAcos0
I'm thinking of this formula as final:
$$\Delta q=\frac{BA}{R}(1-\left| cos\beta \right|)$$
$$\Delta q=\frac{BA}{R}(cos\alpha-cos(\alpha+\beta))$$ .
It is said that in the beginning, the Area was perpendicular to magnetic field lines so the
##\alpha=0##
##\phi_{1}=BAcos0=BA## ,
Refered to that:
$$\Delta q=\frac{BA}{R}(1-cos\beta)$$
but, thinking logically, cosine function becomes negative at 120 degrees so the subtraction of cosines becomes the sum. And I do not think that ##\Delta q## can be bigger than BAcos0
I'm thinking of this formula as final:
$$\Delta q=\frac{BA}{R}(1-\left| cos\beta \right|)$$
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