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?
0.73 Tesla
6.60 Tesla
0.34 Tesla
0.61 Tesla
0.73 Tesla
To find the magnitude of the magnetic field acting on the conductor, we can use the formula for the force on a current-carrying conductor in a magnetic field:
F = BILsinθ
Where:
F is the force on the conductor (given as 2.2 N),
B is the magnitude of the magnetic field,
I is the current flowing through the conductor (given as 1.5A),
L is the length of the conductor (given as 2m),
θ is the angle between the direction of the current and the magnetic field (assumed to be 90 degrees because the magnetic field acts perpendicular to the flow of current).
Rearranging the formula to solve for B, we get:
B = F / (ILsinθ)
Now, we can substitute the given values into the formula:
B = 2.2 N / (1.5A * 2m * sin(90°))
Simplifying further:
B = 2.2 N / (3A * 2m * 1)
B = 0.73 T
Therefore, the magnitude of the magnetic field that acts perpendicular to the flow of current in the conductor is 0.73 Tesla.