A thin 2.82 m long copper rod in a uniform
magnetic field has a mass of 44.2 g. When the
rod carries a current of 0.238 A, it floats in
the magnetic field.
The acceleration of gravity is 9.81 m/s2 .
What is the field strength of the magnetic
field?
Answer in units of T.
7996
To find the field strength of the magnetic field, we can use the principle of magnetic levitation. When a current-carrying conductor is placed in a magnetic field, a magnetic force is exerted on the conductor, which opposes the gravitational force.
The equation that relates the magnetic force, current, length of the conductor, magnetic field strength, and gravitational force is given by:
Fmag = Fgrav
where,
Fmag is the magnetic force,
Fgrav is the gravitational force on the rod,
I is the current,
L is the length of the rod,
B is the magnetic field strength, and
g is the acceleration due to gravity.
The magnetic force, Fmag, is given by the equation:
Fmag = BIL
The gravitational force, Fgrav, is given by the equation:
Fgrav = mg
where,
m is the mass of the rod, and
g is the acceleration due to gravity.
Since the rod is floating in the magnetic field, the magnetic force and the gravitational force are equal. Therefore, we can set the equations for the magnetic force and the gravitational force equal to each other:
BIL = mg
Rearranging the equation solves for B:
B = mg / (IL)
Now we can substitute the given values into the equation to find the field strength, B:
m = 44.2 g = 0.0442 kg (converting grams to kilograms)
I = 0.238 A
L = 2.82 m
g = 9.81 m/s^2
B = (0.0442 kg × 9.81 m/s^2) / (0.238 A × 2.82 m)
B ≈ 0.0804 T
Therefore, the field strength of the magnetic field is approximately 0.0804 Tesla (T).