Question

A rectangular loop of wire is pushed into a region of uniform magnetic field. Given the data in the figure, what is the magni
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Answer #1

The width of the loop is 0.2 m and the length is 0.5 m (actually, the length is not relevant to this problem)

The magnetic field is B = -0.6 T and the velocity is -10 m/s (since it is leftwards) (again, since only the magnitude of the voltage is required, the signs are not important)

The voltage is induced by the change in the flux associated with the loop. Assuming the magnetic field to not have any edge effects and being perpendicular to the loop, the flux that passes through the loop is the magnetic field times the area of the part of the loop that is inside the field.

\phi=BA\implies \frac{d\phi}{dt}=\frac{d(BA)}{dt}=B\frac{dA}{dt}

Since the magnetic field does not change, it comes out of the differentiated part.

The area inside the loop is

l \times b = (l_0+vt) \times 0.2

where l_0 was the initial length that was inside the magnetic field at t=0 s.

Thus, the rate of change of the flux is then:

\frac{d\phi}{dt} = B\frac{d}{dt}((l_0+vt) \times 0.2)=B(0.2v)=-0.6\times 0.2\times 10=-1.2

The induced voltage is given by:

|V|=|\frac{d\phi}{dt}|=|-1.2| = 1.2\text{ V}

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