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
.73
To find the magnitude of the magnetic field acting perpendicular to the flow of current in the conductor, you can use the formula for the magnetic force on a current-carrying conductor, which is given by:
F = BIL
Where:
F is the force on the conductor (2.2 N),
B is the magnitude of the magnetic field,
I is the current flowing through the conductor (1.5 A), and
L is the length of the conductor (2 m).
Rearranging the formula, we can solve for B:
B = F / (IL)
Substituting the given values, we get:
B = 2.2 N / (1.5 A * 2 m)
Calculating this expression, we find:
B = 0.73 Tesla
Therefore, the magnitude of the magnetic field that acts perpendicular to the flow of current in the conductor is 0.73 Tesla.