A square wire loop of side \(0.2\,\mathrm{m}\) moves with a constant speed of \(v=25\,\mathrm{m/s}\) through a region containing a uniform magnetic field of \(B=0.15\,\mathrm{T}\), as shown above left. A graph of the magnetic flux \( \phi \) through the loop as a function of time \(t\) is shown above right. Time \(t=0\) occurs when the right edge of the loop just begins to enter the magnetic field.
What is the total width of the magnetic field region through which the loop moves?
Correct Answer: \( \boxed{\mathrm{C}} \)
The graph shows that the magnetic flux remains constant from \(t=8\,\mathrm{ms}\) to \(t=16\,\mathrm{ms}\). During this interval, the entire loop is completely inside the magnetic field.
The loop travels a distance
\( d_{1}=v\Delta t=(25)(8\times10^{-3})=0.2\,\mathrm{m} \)
This equals the side length of the square loop, confirming the graph.
The total time from the instant the front edge enters until the rear edge leaves the field is
\( 24\,\mathrm{ms}=0.024\,\mathrm{s} \)
Hence, the total distance traveled during this time is
\( d_{\text{total}}=v(0.024)=25(0.024)=0.6\,\mathrm{m} \)
This distance equals the magnetic field width plus one side length of the loop:
\( W+0.2=0.6 \)
Therefore,
\( W=0.6-0.2=0.4\,\mathrm{m} \)
Thus, the width of the magnetic field region is \(0.4\,\mathrm{m}\).
Therefore, the correct answer is (C).
A circular loop of radius \(r\) is located in a uniform magnetic field of magnitude \(B\) directed at an angle \( \theta \) to the plane of the loop, as shown above. What is the magnetic flux through the loop?
Correct Answer: \( \boxed{\mathrm{A}} \)
Magnetic flux through a surface is given by
\( \phi = BA\cos\alpha \)
where \( \alpha \) is the angle between the magnetic field and the normal to the surface.
In this problem, the angle \( \theta \) is measured with respect to the plane of the loop, so the angle between the magnetic field and the normal is
\( \alpha = 90^\circ-\theta \)
Therefore,
\( \phi = BA\cos(90^\circ-\theta)=BA\sin\theta \)
Since the area of the circular loop is
\( A=\pi r^{2} \)
the magnetic flux becomes
\( \boxed{\phi=\pi r^{2}B\sin\theta} \)
Therefore, the correct answer is (A).
A magnetic field of magnitude \(4.0\,\mathrm{T}\) is directed at an angle of \(30^{\circ}\) to the plane of a rectangular loop of area \(5.0\,\mathrm{m^2}\), as shown above. What is the magnetic flux through the loop?
Correct Answer: \( \boxed{\mathrm{A}} \)
Magnetic flux through a surface is given by
\( \Phi = BA\cos\theta \)
where \( \theta \) is the angle between the magnetic field and the normal to the surface.
Since the magnetic field makes an angle of \(30^{\circ}\) with the plane of the loop, the angle with the normal is
\( \theta = 90^{\circ}-30^{\circ}=60^{\circ} \)
Therefore,
\( \Phi = BA\cos60^{\circ} \)
\( \Phi = (4.0\,\mathrm{T})(5.0\,\mathrm{m^2})\left(\dfrac{1}{2}\right) \)
\( \Phi = 10\,\mathrm{T\cdot m^2} \)
Equivalently, because the given angle is measured from the plane of the loop, the flux may also be written as \( \Phi = BA\sin30^{\circ} \), which gives the same result.
Therefore, the correct answer is (A).
