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Assume that the rod in Fig.31.2 has a length of 0.86 m,the resistor has value 2.2 ,and a magnetic field of 8.0 T is directed int

frutty [35]

1) 0.48 m/s

2) Clockwise

3) 10.3 N

Explanation:

1)

The figure is missing: find it in attachment.

As the rod is pulled along the rails, in the region of magnetic field, an electromotive force is induced in the rod itself, due to the change in magnetic flux through the circuit enclosed by the rod and the rails.

The magnitude of this induced electromotive force (which is equivalent to the potential difference induced in the rod) is

$V=BvL$

where:

B is the strength of the magnetic field

v is the speed of the rod

L is the length of the rod

Also we know that according to Ohm's law, the potential difference across the resistance is equal to

$V=RI$

where

R is the resistance of the rod

I is the current

Since the potential difference across the resistor is equal to the potential difference across the rod, we can combine the two equations, and we get:

$BvL=RI\\v=\frac{RI}{BL}$

Here we have:

$R=2.2 \Omega\\B=8.0 T\\L=0.86 m\\I=1.5 A$

Therefore, the speed of the rod must be:

$v=\frac{(2.2)(1.5)}{(8.0)(0.86)}=0.48 m/s$

2)

Here we have to apply Lenz's law, which states that the direction of the induced emf (and so, of the induced current) is such that it opposes the change in magnetic flux through the system.

First of all, here we notice that as the rod moves, the area enclosed by the circuit decreases: this means that the magnetic flux through the circuit decreases too.

As a result, the induced current must produce a magnetic field which goes in the same direction as the external field, in order to "restore" the magnetic flux. Here the external magnetic field points inside the paper, so the magnetic field produced by the induced current must also point inside the paper.

In order for that to happen, we see that the induced current must flow clockwise: in fact, if we take the rod and we apply the right-hand rule, we see that if we place the thumb towards the left (direction of the clockwise current), the other fingers "wrapped" point into the paper at the center of the circuit, so this is the correct direction.

3)

A current-carrying wire in a magnetic field experiences a force of magnitude

$F=ILB$

where

I is the current

L is the length of the wire

B is the strength of the magnetic field

Here we have

I = 1.5 A

L = 0.86 m

B = 8.0 T

So the force experienced by the rod is

$F=(1.5)(8.0)(0.86)=10.3 N$

Therefore, we must apply an equal and opposite force of 10.3 N in order to maintain keep the acceleration of the rod to zero, so that the velocity is constant, and the current is constant as well.