Video Transcript
In a physics experiment that aims
to measure the magnetic field of Earth, an antenna of length 1.2 meters is fixed on
the ceiling of a train that moves in a direction perpendicular to the horizontal
magnetic field lines of Earth. If the velocity of the train is 200
kilometers per hour and the measured emf at the antenna terminals is 1.2 millivolts,
then the horizontal component of Earth’s magnetic field is blank. (A) 5.5 times 10 to the four
tesla. (B) 5.0 times 10 to the negative
six tesla. (C) 1.8 times 10 to the negative
five tesla. (D) 8.0 times 10 to the negative
three tesla.
In this question, we are told about
a physics experiment designed to measure the magnetic field of the Earth. This experiment consists of a
1.2-meter antenna that is fixed onto the ceiling of a train that moves
perpendicularly to the horizontal magnetic field lines of Earth. The train has a velocity of 200
kilometers per hour. We are told that the antenna
measures an emf, which we’ll call 𝜖, of 1.2 millivolts. Given all of this information, we
are asked to figure out what the horizontal component of the Earth’s magnetic field
is. In order to solve this, we will
need to recall some information about emf and how we can calculate it.
When a straight length of a
conductor is moving through a magnetic field, an electromotive force, or emf, is
induced in the conductor. We can describe the emf using the
following equation. Emf, 𝜖, is equal to the length of
the conductor, 𝑙, multiplied by the magnitude of the velocity it is traveling at,
𝑣, multiplied by the strength of the magnetic field, 𝐵, multiplied by the sin of
the angle 𝜃 it forms with the direction of the magnetic field. In this question, we are told that
the antenna is oriented vertically, while the magnetic field lines are
horizontal. This means there is a 90-degree
angle between the antenna and the magnetic field. The sin of a 90-degree angle is
equal to one, so we can go ahead and sub this into the equation. In our case then, the equation
becomes emf, 𝜖, is equal to 𝑙 times 𝑣 times 𝐵.
We’re being asked to find the
strength of the horizontal component of Earth’s magnetic field. That’s the quantity 𝐵 in our
equation. This means we need to rearrange the
equation to make 𝐵 the subject. To do this, we’ll divide both sides
of the equation by both length 𝑙 and velocity 𝑣. The 𝑙 and the 𝑣 then cancel from
the numerator and denominator on the right-hand side. We find, then, that the strength of
the magnetic field is equal to the emf divided by the length of the antenna and the
velocity it is traveling at.
Before we can substitute in our
values, we’ll need to convert them into the correct SI units. The length is already in meters, so
we do not need to convert this. The velocity is given in kilometers
per hour, so we’ll need to convert this to meters per second. We know that there are 1000 meters
in one kilometer. So multiplying 200 kilometers per
hour by 1000 meters per kilometer, we get a velocity of 200,000 meters per hour. Now, there are 60 minutes in one
hour and 60 seconds in one minute. So, we need to divide this velocity
value by 60 minutes per hour and then divide that by 60 seconds per minute to get a
speed in meters per second. Doing this gives us a velocity of
55.5 recurring meters per second.
Finally, we need to convert the
value of the emf from the given units of millivolts into units of volts. Recall that there are 1000
millivolts in one volt or, equivalently, one millivolt is equal to 10 to the
negative three volts. So, to convert from millivolts to
volts, we need to multiply the given value by 10 to the negative three volts per
millivolt. This gives us an emf of 0.0012
volts.
We now have everything that we need
in order to solve this equation for the field strength 𝐵. When we substitute in our values,
we have that 𝐵 is equal to 0.0012 volts divided by 1.2 meters and 55.5 recurring
meters per second. Before evaluating this expression,
we should take a look at the units we have. On the right-hand side, we have
units of volts in the numerator divided by meters multiplied by meters per second in
the denominator. We can write this more simply as
volt seconds per meter squared. Now, units of volt seconds per
meter squared are equivalent to units of tesla, the unit of magnetic field
strength.
Knowing that we have the correct
units, let’s now evaluate this expression for 𝐵. After calculating this, we find
that 𝐵, the strength of the horizontal component of Earth’s magnetic field, is
equal to 1.8 times 10 to the negative five tesla. This result matches the value given
in option (C). We have then that the correct
answer is option (C). The horizontal component of Earth’s
magnetic field is 1.8 times 10 to the negative five tesla.
