A dc generator is rated at $16 \mathrm{~kW}, 250 \mathrm{~V}$, and $56 \mathrm{~A}$ when it rotates at $1200 \mathrm{rpm}$. The resistance of the armature windings is $0.40~ \Omega$. Calculate the "no-load" voltage at $1200 \mathrm{~rpm}$ (when there is no circuit hooked up to the generator).

A dc generator is rated at $16 \mathrm{~kW}, 250 \mathrm{~V}$, and $64 \mathrm{~A}$ when it rotates at $1000 \mathrm{rpm}$. The resistance of the armature windings is $0.40 \Omega$. (a) Calculate the "no-load" voltage at $1000 \mathrm{rpm}$ (when there is no circuit hooked up to the generator).

A dc generator is rated at $16 \mathrm{~kW}, 250 \mathrm{~V}$, and $64 \mathrm{~A}$ when it rotates at $1000 \mathrm{rpm}$. The resistance of the armature windings is $0.40 \Omega$. (b) Calculate the full-load voltage (i.e. at $64 \mathrm{~A}$ ) when the generator is run at $750 \mathrm{rpm}$. Assume that the magnitude of the magnetic field remains constant.

In a DC machine, the air gap flux density is $4 \mathrm{~Wb} / \mathrm{m}^2$. The area of the pole face is $2 \mathrm{~cm} \times 4 \mathrm{~cm}$. Find the flux per pole in the machine.

Calculate the peak output voltage of a simple generator whose square armature windings are 6.60 cm on a side; the armature contains 155 loops and rotates in a field of 0.200 T at a rate of 120 rev/s.

The magnetic field perpendicular to a single 13.2-cm-diameter circular loop of copper wire decreases uniformly from 0.750 T to zero. If the wire is 2.25 mm in diameter, how much charge moves past a point in the coil during this operation?

A circular loop in the plane of the paper lies in a $0.65~\mathrm{T}$ uniform magnetic field pointing into the paper. The loop's diameter changes from $20.0 \mathrm{~cm}$ to $8.0 \mathrm{~cm}$ in $0.50 \mathrm{~s}$. What is the average induced current if the coil resistance is $2.5 \Omega$ ?

A circular loop in the plane of the paper lies in a $0.65~\mathrm{T}$ uniform magnetic field pointing into the paper. The loop's diameter changes from $20.0 \mathrm{~cm}$ to $8.0 \mathrm{~cm}$ in $0.50 \mathrm{~s}$. What is the magnitude of the average induced emf?

Suppose that the U-shaped conductor and connecting rod in Fig. 29-12a are oriented vertically (but still in contact) so that the rod is falling due to the gravitational force. Find the terminal speed of the rod if it has mass $m=3.6$ grams, length $\ell=18 \mathrm{~cm}$, and resistance $R=0.0013 \Omega$. It is falling in a uniform horizontal field $B=0.060 \mathrm{~T}$. Neglect the resistance of the U-shaped conductor.

A circular loop in the plane of the paper lies in a $0.65~\mathrm{T}$ uniform magnetic field pointing into the paper. The loop's diameter changes from $20.0 \mathrm{~cm}$ to $8.0 \mathrm{~cm}$ in $0.50 \mathrm{~s}$. What is the direction of the induced current?

In Fig. 29-12a, the rod moves to the right with a speed of $1.3 \mathrm{~m} / \mathrm{s}$ and has a resistance of $2.5 \Omega$. The rail separation is $\ell=25.0 \mathrm{~cm}$. The magnetic field is $0.35 \mathrm{~T}$, and the resistance of the U-shaped conductor is $25.0 \Omega$ at a given instant. Calculate (a) the induced emf.

The magnetic flux through each loop of a 75 -loop coil is given by $\left(8.8 t-0.51 t^3\right) \times 10^{-2} \mathrm{~T} \cdot \mathrm{m}^2$, where the time $t$ is in seconds (a) Determine the emf $\mathscr{E}$ as a function of time.

The magnetic flux through each loop of a $75$ - loop coil is given by $\left(9.6 t-0.51 t^3\right) \times 10^{-2} \mathrm{~T} \cdot \mathrm{m}^2$, where the time $t$ is in seconds. $(a)$ Determine the emf $\mathscr{E}$ as a function of time. $(b)$ What is $\mathscr{E}$ at $t=1.0 \mathrm{~s}$ and $t=4.0 \mathrm{~s}$ ?

The magnetic field perpendicular to a single $13.2~\mathrm{cm}$ diameter circular loop of copper wire decreases uniformly from $0.760 \mathrm{~T}$ to zero. If the wire is $2.25 \mathrm{~mm}$ in diameter, how much charge moves past a point in the coil during this operation?