A current-carrying conductor that is 8.00x10^-2 m long is placed perpendicular to a magnetic field that has a strength of 6.00x10^-3 T. During a 3.00s time interval, 7.50x10^19 electrons pass a point in the conductor. The magnitude of the average magnetic force exerted on the conductor in scientific notation is a.bcx10^-d N ?
a. 2.25
b. 4.50
c. 6.75
d. 9.00
Answer: b. 4.50x10^-3 N
To find the magnitude of the average magnetic force exerted on the conductor, we can use the equation:
F = BIL sin(θ)
Where:
F is the magnetic force,
B is the magnetic field strength,
I is the current flowing through the conductor,
L is the length of the conductor, and
θ is the angle between the magnetic field and the current.
Since the conductor is placed perpendicular to the magnetic field, the angle (θ) between them is 90 degrees.
Given:
L = 8.00 × 10^-2 m
B = 6.00 × 10^-3 T
I = (number of electrons passing a point in a second) × (charge of each electron)
First, let's calculate the current (I):
The number of electrons passing through the conductor in 3 seconds is:
n = 7.50 × 10^19 electrons
Since 1 Ampere (A) is equivalent to 1 Coulomb (C) passing through a point in 1 second, we need to convert the number of electrons to Coulombs:
1 electron = 1.6 × 10^-19 C
Therefore, the total charge passing through the conductor is:
q = n × charge of each electron
= (7.50 × 10^19 electrons) × (1.6 × 10^-19 C)
= 1.20 C
Since the time interval is 3 seconds:
I = q / t
= 1.20 C / 3 s
= 0.40 A
Now we can calculate the magnetic force (F):
F = BIL sin(θ)
= (6.00 × 10^-3 T) × (0.40 A) × sin(90°)
= (6.00 × 10^-3 T) × (0.40 A) × 1
= 2.40 × 10^-3 N
Therefore, the magnitude of the average magnetic force exerted on the conductor is 2.40x10^-3 N.
To find the magnitude of the average magnetic force exerted on the conductor, you can use the formula F = (BIL) sin(theta), where F is the force, B is the magnetic field strength, I is the current, L is the length of the conductor, and theta is the angle between the magnetic field and the conductor.
First, let's find the current. We know that 7.50x10^19 electrons pass through a point in the conductor during a time interval of 3.00s. The current (I) is given by the formula I = Q/t, where Q is the total charge passing through the conductor and t is the time interval. Each electron has a charge of 1.6x10^-19 C, so the total charge passing through the conductor is 7.50x10^19 electrons * 1.6x10^-19 C/electron = 12 C. Therefore, the current is I = 12 C / 3.00s = 4 A.
Next, let's calculate the force. We know that the length of the conductor (L) is 8.00x10^-2m and the magnetic field strength (B) is 6.00x10^-3T. The angle between the magnetic field and the conductor is 90 degrees since the conductor is perpendicular to the magnetic field (sin(theta) = sin(90) = 1).
Now we can substitute the values into the formula F = (BIL) sin(theta):
F = (6.00x10^-3T)(4A)(8.00x10^-2m)(1)
F = 1.92x10^-4 N
So, the magnitude of the average magnetic force exerted on the conductor is 1.92x10^-4 N in scientific notation.