201. A fixed \(150\)-turn coil of area \(4.0\times10^{-3}\,m^2\) is placed with its area vector parallel to a magnetic field. The magnetic field decreases uniformly from \(0.80\,T\) to \(0.20\,T\) in \(0.30\,s\). The average induced emf is
ⓐ. \(+1.2\,V\)
ⓑ. \(-1.2\,V\)
ⓒ. \(+0.12\,V\)
ⓓ. \(-0.12\,V\)
Correct Answer: \(+1.2\,V\)
Explanation: \( \textbf{Given:} \) \(N=150\), \(A=4.0\times10^{-3}\,m^2\), \(B_1=0.80\,T\), \(B_2=0.20\,T\), and \(\Delta t=0.30\,s\).
\( \textbf{Orientation:} \) \(\vec{A}\parallel\vec{B}\), so \(\cos\theta=1\).
\( \textbf{Change in magnetic field:} \)
\[
\Delta B=B_2-B_1
\]
\[
\Delta B=0.20-0.80=-0.60\,T
\]
\( \textbf{Flux change per turn:} \)
\[
\Delta\phi_B=A\Delta B
\]
\[
\Delta\phi_B=(4.0\times10^{-3})(-0.60)=-2.4\times10^{-3}\,Wb
\]
\( \textbf{Average Faraday law:} \)
\[
\varepsilon_{avg}=-N\frac{\Delta\phi_B}{\Delta t}
\]
\( \textbf{Substitution:} \)
\[
\varepsilon_{avg}=-150\left(\frac{-2.4\times10^{-3}}{0.30}\right)
\]
\( \textbf{Calculation:} \)
\[
\varepsilon_{avg}=+1.2\,V
\]
\( \textbf{Final answer:} \) The average induced emf is \(+1.2\,V\).
202. A fixed coil is kept in a magnetic field whose direction is perpendicular to the coil plane. The \(B\)-\(t\) graph has a steeper slope in interval P than in interval Q. If the area and number of turns are unchanged, the induced emf magnitude is
ⓐ. zero in both intervals
ⓑ. larger in interval P
ⓒ. independent of the slope of the \(B\)-\(t\) graph
ⓓ. larger in interval Q
Correct Answer: larger in interval P
Explanation: For a fixed coil in a field perpendicular to its plane, \(|\varepsilon|=NA\left|\frac{dB}{dt}\right|\). The slope of the \(B\)-\(t\) graph represents \(\frac{dB}{dt}\). A steeper slope means a larger rate of change of magnetic field. Since \(N\) and \(A\) are unchanged, the interval with the larger slope magnitude has the larger induced emf magnitude. The height of the graph alone is less important than how quickly it changes.
203. A fixed coil of resistance \(4.0\,\Omega\) has \(100\) turns, each of area \(1.5\times10^{-3}\,m^2\). Its area vector is parallel to a magnetic field increasing at \(0.80\,T\,s^{-1}\). The induced current magnitude is
ⓐ. \(3.0\times10^{-2}\,A\)
ⓑ. \(1.9\times10^{-3}\,A\)
ⓒ. \(1.2\times10^{-1}\,A\)
ⓓ. \(4.8\times10^{-1}\,A\)
Correct Answer: \(3.0\times10^{-2}\,A\)
Explanation: \( \textbf{Given:} \) \(N=100\), \(A=1.5\times10^{-3}\,m^2\), \(\left|\frac{dB}{dt}\right|=0.80\,T\,s^{-1}\), and \(R=4.0\,\Omega\).
\( \textbf{Orientation:} \) \(\vec{A}\parallel\vec{B}\), so \(\cos\theta=1\).
\( \textbf{Emf magnitude:} \)
\[
|\varepsilon|=NA\left|\frac{dB}{dt}\right|
\]
\( \textbf{Substitution:} \)
\[
|\varepsilon|=100(1.5\times10^{-3})(0.80)
\]
\( \textbf{Emf calculation:} \)
\[
|\varepsilon|=0.12\,V
\]
\( \textbf{Current relation:} \)
\[
I=\frac{|\varepsilon|}{R}
\]
\( \textbf{Substitution for current:} \)
\[
I=\frac{0.12}{4.0}
\]
\( \textbf{Final answer:} \) The induced current magnitude is \(3.0\times10^{-2}\,A\).
204. A fixed loop is placed in a magnetic field whose magnitude varies with time. The area vector of the loop is perpendicular to \(\vec{B}\) throughout. The induced emf due to this varying \(B\) is zero because
ⓐ. the magnetic field is changing too slowly
ⓑ. resistance cancels the magnetic flux
ⓒ. the field has no component along \(\vec{A}\)
ⓓ. the loop is made of conducting material
Correct Answer: the field has no component along \(\vec{A}\)
Explanation: Magnetic flux through a plane surface is \(\phi_B=BA\cos\theta\). If \(\vec{B}\perp\vec{A}\), then \(\theta=90^\circ\) and \(\cos90^\circ=0\). The flux through the loop is therefore zero at all times for that orientation. Even if \(B\) changes in magnitude, the product \(BA\cos90^\circ\) remains zero. A changing field produces induction only when it changes the flux linked with the loop.
205. A table describes a fixed loop in a time-varying magnetic field.
| Row | Condition | Conclusion |
|---|
| P | \(\vec{A}\parallel\vec{B}\), \(B\) increasing | Flux increases |
| Q | \(\vec{A}\parallel\vec{B}\), \(B\) constant | No flux change |
| R | \(\vec{A}\perp\vec{B}\), \(B\) changing | Flux remains zero |
| S | \(\vec{A}\parallel\vec{B}\), \(B\) decreasing | Flux must increase |
The row with the faulty conclusion is
ⓐ. Row Q
ⓑ. Row R
ⓒ. Row S
ⓓ. Row P
Correct Answer: Row S
Explanation: When \(\vec{A}\parallel\vec{B}\), the flux is \(\phi_B=BA\). If \(B\) increases, flux increases; if \(B\) is constant, flux is constant. When \(\vec{A}\perp\vec{B}\), \(\phi_B=0\), so changing \(B\) does not change the flux as long as the orientation stays the same. Row S is faulty because with \(\vec{A}\parallel\vec{B}\), decreasing \(B\) makes the flux decrease, not increase. The sign and direction can be handled later, but the magnitude trend follows \(B\).
206. A rectangular coil rotates uniformly in a uniform magnetic field. The magnetic flux linked with it varies with time because
ⓐ. the \(\vec{B}\)-area-vector angle changes continuously
ⓑ. the resistance of the coil changes into magnetic flux
ⓒ. the magnetic field must disappear every half turn
ⓓ. the area of the rigid coil becomes zero twice in each rotation
Correct Answer: the \(\vec{B}\)-area-vector angle changes continuously
Explanation: A rigid rotating coil has fixed area. If it rotates in a uniform magnetic field, the field strength \(B\) may also remain constant. The changing quantity is the angle between \(\vec{B}\) and the coil's area vector \(\vec{A}\). Since flux depends on \(\phi_B=BA\cos\theta\), the changing angle makes the flux vary with time. This angular variation is the basis of sinusoidal flux in a rotating coil.
207. For a rotating \(N\)-turn coil in a uniform magnetic field, the magnetic flux linkage is commonly written as
ⓐ. \(N\phi_B=NBA+\omega t\)
ⓑ. \(N\phi_B=NBA\cos\omega t\)
ⓒ. \(N\phi_B=\frac{N\omega t}{BA}\)
ⓓ. \(N\phi_B=RBA\sin\omega t\)
Correct Answer: \(N\phi_B=NBA\cos\omega t\)
Explanation: For one turn of a rotating coil, the flux is \(\phi_B=BA\cos\theta\). In uniform rotation, the angle can be written as \(\theta=\omega t\) when the coil starts with its area vector parallel to \(\vec{B}\). For \(N\) turns, flux linkage becomes \(N\phi_B=NBA\cos\omega t\). The expression is sinusoidal because \(\cos\omega t\) varies periodically with time. Resistance is not part of the flux-linkage expression.
208. At the instant when the area vector of a rotating coil is parallel to \(\vec{B}\), the magnetic flux through the coil is
ⓐ. maximum
ⓑ. always negative
ⓒ. independent of area
ⓓ. zero
Correct Answer: maximum
Explanation: Magnetic flux through a plane coil is \(\phi_B=BA\cos\theta\). When the area vector is parallel to \(\vec{B}\), \(\theta=0^\circ\). Since \(\cos0^\circ=1\), the flux has maximum magnitude \(BA\) for one turn. For an \(N\)-turn coil, the corresponding maximum flux linkage is \(NBA\). This is a maximum-flux position, not necessarily a maximum-emf position.
209. At the instant when the plane of a rotating coil is parallel to the magnetic field, the magnetic flux through the coil is
ⓐ. equal to \(NBA\) for one turn
ⓑ. maximum positive
ⓒ. maximum negative
ⓓ. zero
Correct Answer: zero
Explanation: The area vector of a coil is perpendicular to the plane of the coil. If the plane of the coil is parallel to the magnetic field, then the area vector is perpendicular to \(\vec{B}\). The angle between \(\vec{B}\) and \(\vec{A}\) is \(90^\circ\). Therefore \(\phi_B=BA\cos90^\circ=0\) for one turn. This position is important because the flux is zero but is changing fastest during uniform rotation.
210. For a rotating coil with flux linkage \(N\phi_B=NBA\cos\omega t\), the induced emf is
ⓐ. \(\varepsilon=NBA\cos\omega t\)
ⓑ. \(\varepsilon=\frac{NBA}{\omega}\sin\omega t\)
ⓒ. \(\varepsilon=NBA\omega t\)
ⓓ. \(\varepsilon=NBA\omega\sin\omega t\)
Correct Answer: \(\varepsilon=NBA\omega\sin\omega t\)
Explanation: \( \textbf{Flux linkage:} \)
\[
N\phi_B=NBA\cos\omega t
\]
\( \textbf{Faraday's law:} \)
\[
\varepsilon=-\frac{d(N\phi_B)}{dt}
\]
\( \textbf{Differentiate the flux linkage:} \)
\[
\frac{d}{dt}(NBA\cos\omega t)=-NBA\omega\sin\omega t
\]
\( \textbf{Apply the negative sign:} \)
\[
\varepsilon=-(-NBA\omega\sin\omega t)
\]
\( \textbf{Simplification:} \)
\[
\varepsilon=NBA\omega\sin\omega t
\]
\( \textbf{Final answer:} \) The induced emf is \(\varepsilon=NBA\omega\sin\omega t\).
The extra factor \(\omega\) appears because the time derivative of \(\cos\omega t\) is involved.
211. In a rotating coil, magnetic flux is maximum at a certain instant. At that same instant, the induced emf is
ⓐ. equal to \(NBA\)
ⓑ. independent of angular speed only at that instant
ⓒ. zero
ⓓ. maximum
Correct Answer: zero
Explanation: In uniform rotation, the flux linkage can be written as \(NBA\cos\omega t\). The flux is maximum when \(\cos\omega t=\pm1\). At these instants, the sine term in \(\varepsilon=NBA\omega\sin\omega t\) is zero. Therefore the induced emf is zero when the flux is maximum or minimum. Emf depends on the rate of change of flux, and the flux has zero instantaneous rate of change at its extreme values.
212. In a uniformly rotating coil, the induced emf is maximum when
ⓐ. the angular speed is zero
ⓑ. magnetic flux through the coil is maximum
ⓒ. the coil has no area
ⓓ. magnetic flux through the coil is zero
Correct Answer: magnetic flux through the coil is zero
Explanation: For a rotating coil, the flux varies as \(\phi_B\propto\cos\omega t\), while the induced emf varies as \(\varepsilon\propto\sin\omega t\). The sine term has maximum magnitude when the cosine term is zero. Physically, this is the position where the flux is changing most rapidly. Maximum flux itself gives zero emf because the flux is momentarily turning around from increasing to decreasing or from decreasing to increasing. The maximum-emf position is therefore a fastest-flux-change position.
213. For a rotating coil with \(N=200\), \(A=5.0\times10^{-3}\,m^2\), \(B=0.40\,T\), and \(\omega=50\,rad\,s^{-1}\), the peak induced emf is
ⓐ. \(10\,V\)
ⓑ. \(5.0\,V\)
ⓒ. \(2.0\,V\)
ⓓ. \(20\,V\)
Correct Answer: \(20\,V\)
Explanation: \( \textbf{Given:} \) \(N=200\), \(A=5.0\times10^{-3}\,m^2\), \(B=0.40\,T\), and \(\omega=50\,rad\,s^{-1}\).
\( \textbf{Peak emf relation:} \)
\[
\varepsilon_0=NBA\omega
\]
\( \textbf{Why this applies:} \) The maximum value of \(\sin\omega t\) is \(1\), so the peak value of \(NBA\omega\sin\omega t\) is \(NBA\omega\).
\( \textbf{Substitution:} \)
\[
\varepsilon_0=(200)(0.40)(5.0\times10^{-3})(50)
\]
\( \textbf{Intermediate multiplication:} \)
\[
(200)(5.0\times10^{-3})=1.0
\]
\[
(1.0)(0.40)(50)=20
\]
\( \textbf{Final answer:} \) The peak induced emf is \(20\,V\).
214. A coil rotates uniformly in a magnetic field. If \(N\), \(B\), and \(A\) are unchanged but the angular speed is doubled, the peak induced emf
ⓐ. becomes four times
ⓑ. becomes double
ⓒ. remains unchanged
ⓓ. becomes half
Correct Answer: becomes double
Explanation: The peak induced emf of a rotating coil is \(\varepsilon_0=NBA\omega\). If \(N\), \(B\), and \(A\) are constant, then \(\varepsilon_0\) is directly proportional to \(\omega\). Doubling \(\omega\) doubles the maximum rate at which flux changes. The maximum flux \(NBA\) itself does not double because it does not contain \(\omega\). Faster rotation increases peak emf by increasing the rate of flux change, not by increasing the maximum flux.
215. For the rotating-coil relation \(N\phi_B=NBA\cos\omega t\), when \(\omega t=90^\circ\), the flux linkage and induced emf are respectively
ⓐ. \(0\) and \(0\)
ⓑ. maximum in magnitude and \(0\)
ⓒ. \(0\) and maximum in magnitude
ⓓ. \(NBA\) and \(NBA\omega\)
Correct Answer: \(0\) and maximum in magnitude
Explanation: For the rotating coil, the flux linkage is \(N\phi_B=NBA\cos\omega t\). At \(\omega t=90^\circ\), \(\cos90^\circ=0\), so the flux linkage is zero. The induced emf is \(\varepsilon=NBA\omega\sin\omega t\). At the same instant, \(\sin90^\circ=1\), so the emf has maximum positive value \(NBA\omega\). This shows that maximum emf occurs when the flux is crossing through zero, not when the flux is maximum.
216. For a coil rotating with \(N=100\), \(B=0.50\,T\), \(A=2.0\times10^{-3}\,m^2\), and \(\omega=100\,rad\,s^{-1}\), the induced emf at an instant when \(\sin\omega t=0.60\) is
ⓐ. \(6.0\,V\)
ⓑ. \(10\,V\)
ⓒ. \(3.0\,V\)
ⓓ. \(12\,V\)
Correct Answer: \(6.0\,V\)
Explanation: \( \textbf{Given:} \) \(N=100\), \(B=0.50\,T\), \(A=2.0\times10^{-3}\,m^2\), \(\omega=100\,rad\,s^{-1}\), and \(\sin\omega t=0.60\).
\( \textbf{Emf equation:} \)
\[
\varepsilon=NBA\omega\sin\omega t
\]
\( \textbf{Peak factor first:} \)
\[
NBA\omega=(100)(0.50)(2.0\times10^{-3})(100)
\]
\( \textbf{Intermediate multiplication:} \)
\[
(100)(2.0\times10^{-3})=0.20
\]
\[
(0.20)(0.50)(100)=10
\]
\( \textbf{Instantaneous emf:} \)
\[
\varepsilon=(10)(0.60)
\]
\( \textbf{Final answer:} \) The induced emf is \(6.0\,V\).
217. In a rotating coil, doubling the angular speed \(\omega\) while keeping \(N\), \(B\), and \(A\) unchanged affects the maximum flux linkage and peak emf as
ⓐ. both maximum flux linkage and peak emf doubled
ⓑ. maximum flux linkage unchanged, peak emf doubled
ⓒ. both maximum flux linkage and peak emf unchanged
ⓓ. maximum flux linkage doubled, peak emf unchanged
Correct Answer: maximum flux linkage unchanged, peak emf doubled
Explanation: The maximum flux linkage of a rotating coil is \(NBA\). It depends on the number of turns, magnetic field, and area, but not on angular speed. The peak induced emf is \(\varepsilon_0=NBA\omega\). Therefore doubling \(\omega\) doubles the peak emf. Faster rotation does not increase the maximum amount of flux linked with the coil; it increases how rapidly that flux changes with time.
218. Study the table for a uniformly rotating coil.
| Row | Position of coil | Flux | Induced emf |
|---|
| P | Area vector parallel to \(\vec{B}\) | Maximum magnitude | Zero |
| Q | Area vector perpendicular to \(\vec{B}\) | Maximum magnitude | Maximum magnitude |
| R | Plane of coil parallel to \(\vec{B}\) | Maximum magnitude | Zero |
| S | Flux is maximum | Zero | Maximum magnitude |
The row that correctly describes the rotating coil is
ⓐ. Row S
ⓑ. Row Q
ⓒ. Row P
ⓓ. Row R
Correct Answer: Row P
Explanation: When the area vector is parallel to \(\vec{B}\), the flux has maximum magnitude because \(\phi_B=BA\cos0^\circ=BA\). At that instant the flux is at an extreme value, so its instantaneous rate of change is zero. Since induced emf depends on rate of change of flux, the emf is zero there. When the area vector is perpendicular to \(\vec{B}\), the flux is zero but the emf has maximum magnitude. Row P is the only row that keeps the flux-emf phase relation correct.
219. Use the graph description below.
For a rotating coil, the magnetic flux linkage graph is a cosine curve starting from its positive maximum at \(t=0\).
The induced emf graph begins from
ⓐ. positive maximum at \(t=0\)
ⓑ. negative maximum at \(t=0\)
ⓒ. a constant non-zero value for all \(t\)
ⓓ. zero, then positive for small \(t\)
Correct Answer: zero, then positive for small \(t\)
Explanation: If the flux linkage is \(NBA\cos\omega t\), it starts from its positive maximum at \(t=0\). Faraday's law gives \(\varepsilon=-\frac{d}{dt}(NBA\cos\omega t)=NBA\omega\sin\omega t\). At \(t=0\), \(\sin0=0\), so the emf begins from zero. For small positive \(t\), \(\sin\omega t\) is positive, so the induced emf becomes positive. The emf graph is shifted in phase because it depends on the rate of change of flux linkage.
220. The peak flux linkage of a rotating coil is \(0.040\,Wb\), and its angular speed is \(25\,rad\,s^{-1}\). The peak induced emf is
ⓐ. \(25\,V\)
ⓑ. \(1.0\,V\)
ⓒ. \(625\,V\)
ⓓ. \(0.040\,V\)
Correct Answer: \(1.0\,V\)
Explanation: \( \textbf{Peak flux linkage:} \)
\[
(N\phi_B)_0=0.040\,Wb
\]
\( \textbf{Angular speed:} \)
\[
\omega=25\,rad\,s^{-1}
\]
\( \textbf{Flux-linkage form:} \)
\[
N\phi_B=(N\phi_B)_0\cos\omega t
\]
\( \textbf{Peak emf relation:} \)
\[
\varepsilon_0=(N\phi_B)_0\omega
\]
\( \textbf{Substitution:} \)
\[
\varepsilon_0=(0.040)(25)
\]
\( \textbf{Calculation:} \)
\[
\varepsilon_0=1.0\,V
\]
\( \textbf{Final answer:} \) The peak induced emf is \(1.0\,V\).
