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
F=ILBsinα
sinα=1
B=F/IL = 2.2/2•1.5 =0.73 T
To find the magnitude of the magnetic field that acts perpendicular to the flow of current in the conductor, you can use the formula for the magnetic force on a current-carrying conductor.
The formula is given by:
F = BIL
Where:
F is the force experienced by the conductor (given as 2.2 N)
B is the magnetic field
I is the current flowing through the conductor (given as 1.5 A)
L is the length of the conductor (given as 2 m)
Rearrange the formula to solve for B:
B = F / (IL)
Substitute the given values into the formula:
B = 2.2 N / (1.5 A × 2 m)
B = 2.2 N / 3 A.m
B ≈ 0.73 Tesla
Therefore, the magnitude of the magnetic field that acts perpendicular to the flow of current in the conductor is approximately 0.73 Tesla.