A conductor of length 2m carrying a 1.5A current experiences a force of 2.2 N. What is the magnitude of the magnetic field that acts perpendicular to the flow of current in the conductor?
To find the magnitude of the magnetic field acting on the conductor, you can apply the formula for the magnetic force experienced by a current-carrying conductor in a magnetic field.
The formula for the magnetic force (F) on a current-carrying conductor in a magnetic field is given by:
F = B * L * I * sin(θ)
Where:
F is the force experienced by the conductor,
B is the magnitude of the magnetic field,
L is the length of the conductor,
I is the current flowing through the conductor,
and θ is the angle between the direction of the current and the magnetic field.
In this problem, we are given:
L = 2m (length of the conductor),
I = 1.5A (current flowing through the conductor),
F = 2.2N (force experienced by the conductor).
Since the conductor is perpendicular to the magnetic field, the angle (θ) between the direction of the current and the magnetic field is 90 degrees (sin(θ) = 1).
Using the formula mentioned earlier, we can rearrange it to solve for the magnetic field (B):
B = F / (L * I * sin(θ))
Substituting the given values:
B = 2.2N / (2m * 1.5A * 1)
Calculating the result, we find:
B ≈ 0.73333 Tesla
Therefore, the magnitude of the magnetic field acting perpendicular to the flow of current in the conductor is approximately 0.73333 Tesla.