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 perpendicular to the flow of current in the conductor, you can use the formula:
Force (F) = Magnetic Field (B) × Current (I) × Length (L)
In this case, the force (F) is given as 2.2 N, the current (I) is 1.5 A, and the length (L) is 2 m.
Rearranging the formula, we get:
Magnetic Field (B) = Force (F) / (Current (I) × Length (L))
Substituting the given values, we have:
B = 2.2 N / (1.5 A × 2 m)
Simplifying the expression, we get:
B = 2.2 N / 3 A⋅m
Now, divide 2.2 N by 3 A⋅m to get the magnitude of the magnetic field:
B ≈ 0.733 T
Therefore, the magnitude of the magnetic field acting perpendicular to the flow of current in the conductor is approximately 0.733 Tesla.