Download PH 202-1D SI Session 3 Induced EMF and Magnetic Flux A metal

PH 202-1D SI
Session 3
Induced EMF and Magnetic Flux
1. A metal bar, with length 80. cm is falls from an arbitrary height. Once it reaches a
velocity of 22 m/s, there is potential difference between its east and west ends of 6.5 x
10-4 V. Assuming the bar is dropping so that its ends pointed in the East-West direction,
and that the magnetic field points North, calculate the magnitude of the magnetic field
and the polarity of the voltage.
Use EMF=vBL
B= 3.7x10-5 T
Use RHR to determine polarity
(+) is on the East end
2.
Consider the above diagram. L is the vertical length of the rod, B is the magnetic field
magnitude, and the object to the top left is a light bulb, completing the circuit. What is the
shortest distance along the rails that the rod needs to travel for the bulb to remain lit for
0.5 seconds?
Using Ohm’s Law: V=IR
Induced EMF=vBL
v=x/t
Power=P=IV or V2/R, V=sqrt(P*R)
Combining these expressions: sqrt(P*R)=x/t)BL
120 V=(x/0.5 s)(0.40 T)(0.60 m)
x=250 m
Remember, 0.5 seconds was substituted for the time interval because the light bulb
will only stay lit for as long as the rod continues to induce an emf and drive current.
3.
Consider the above diagram. The two surfaces shown have identical areas, A. A uniform
magnetic field, parallel to the yz-plane, exists in the space between the two surfaces.
Calculate the ratio of xz/xy.
xz=BAcos55
xy=BAcos35
When we divide the top equation by the bottom, the two BA terms cancel, and we
are left with cos55/cos35. You can also realize than cos55/cos35 is the same as tan35.
xz/xy=0.70
4. Two conducting rails are in the same plane and make an angle of 19° with respect to each
other. Consider the scenario of a metal rod being placed on top of the rails and sliding
along the rails, beginning at the vertex at a constant velocity of v=0.60 m/s. There is a
0.38 T uniform magnetic field perpendicular to and into the plane of the two rods. What
is the average emf induced in the triangle formed after 6.0 s?
Remember that velocity=displacement/time interval
Displacement of rod= (0.60 m/s)(6.0 s) = 3.6 m
This means than tan(19°)=y/3.6
y=1.24 m is the length of the rod for the triangle at t=6.0 s
Area of the triangle, then: A=1/2(3.6 m)(1.2 m)=2.2 m2
EMFavg= t=(0.836 Wb)/(6.0 s) = 0.14 V
5. Two 0.68 m-long conducting rods are perpendicular to a 4.7 T uniform magnetic field.
The two rods are rotating at the same speed and in opposite directions; each with one end
fixed in place-at the same electric potential. The fixed ends of the rods are connected by a
wire. As the rods rotate, they come to within 1.0 mm of each other, when they are both
perfectly horizontal. A potential difference of 4.5x103 V is required to cause a 1.0 mm
spark in the air. What is the angular speed (in rad/s) of the rods when a spark jumps
across the gap?
The rods will both trace circular paths with radii equal to their lengths.
r=0.68 m
When a spark jumps across the gap, the velocity of the rods in the magnetic field
will have induced an emf of 4.5x103 V
EMF=vBL
v=1410 m/s
The problem wants the angular velocity, v=rv/r
=2100 rad/s