**Problem 1**

Using what we learned in Physics 121, we can determine that the maximum force that can be applied to rod without causing any motion is equal to

The magnetic force acting on the rod is equal to

We thus conclude that the maximum magnetic field for which the rod remains at rest is equal to

**Problem 2**

When the current reaches a stable value (independent of time), there will be no potential drop across the solenoid, the current will be equal to

The magnetic field inside the solenoid, assumed to be perfect, is equal to

**Problem 3**

Using the techniques we discussed in class (see equation 33.3 in the lecture notes) we can conclude that

**Problem 4**

The atomic mass of lithium is 6.941 g. The number of atoms
in 6.941 g of Li is equal to N_{A}. The magnetic moment
of one Li atom is equal to 9.27 x 10^{-24} A m^{2}
(see Table 33.1 in Ohanian). Assuming that all atoms in the sample
are lined up, the total dipole moment of the sample is equal to
the product of the number of atoms times the magnetic moment of
each atom:

The magnetic field produced by this dipole is equal to

**Problem 5**

The magnetic flux intercepted by the coil is equal to

where N is the number of turns, w and h are the width and height of the turns, B is the magnetic field, and [theta] is the angle between the normal of the coil and the magnetic field. The flux will be time dependent since the angle [theta] will be time dependent. Assuming a constant rate of rotation, we can rewrite this equation as

The induced emf is equal to

The maximum emf is thus equal to

The angular frequency required to achieve this emf is equal to

**Problem 6**

During one complete revolution, the rod sweeps an area of

Assuming the time required to complete one revolution is T, we conclude that the rate of flux change is equal to

The magnitude of the induced emf is thus equal to

**Problem 7**

The magnetic flux through the loop is equal to

The induced emf is equal to

The problem states that in a time interval [Delta]t the magnetic field changes from B to fB. The change in the magnetic field, [Delta]B, is thus equal to (f - 1)B. The induced emf is thus equal to

**Problem 8**

The current induced in the loop is equal to

**Problem 9**

Define t = 0 to be the time just before the loop is being pulled. The magnetic flux though the loop at this time is equal to

The flux at time t, when the loop is pulled out of the magnetic field region, is equal to 0. Thus, the rate of the change of flux is equal to

The induced emf is equal to

**Problem 10**

The magnetic field generated by a current-carrying wire is equal to

Consider a small time interval dt. During this time interval the rod moves a distance v dt. The flux that is swept by the rod during this time interval is equal to

The rate of flux change is thus equal to

The magnitude of the induced emf is equal to

**Problem 11**

Consider the ring at time t. The enclosed flux at this time is equal to

The induced emf is equal to

The total resistance of the loop at time t is equal to

The induced current is thus equal to

**Problem 12**

The change in flux that is enclosed by the circuit during a time interval dt is equal to

where L is the distance between the rails, B is the magnetic field, and v is the velocity of the bar. The induced emf in the bar is equal to

The magnitude of the induced current is equal to

The magnetic force acting on the rod as it moved through the magnetic field is equal to

Since the rod is moving with constant velocity v, the net force on the rod must be zero, and the external applied force must be equal in magnitude, but opposite in direction to the magnetic force:

**Problem 13**

The energy dissipated in the current loop is equal to

We conclude that the work done by the external force is converted into Joule heating in the circuit.