A 0.050-m-long wire is carrying a current of 2.0 A at right angles to a 0.15-T magnetic field. What is the magnitude of the force exerted by the magnetic field on the wire?
please help asap
dF = I dL cross B
they are perpendicular so
F = I L B
= 2.0 * 0.050 * 0.15 Newtons
By the way, this is surely in your text.
To find the magnitude of the force exerted by the magnetic field on the wire, you can use the formula for the magnetic force on a current-carrying wire in a magnetic field:
F = I * B * L * sin(theta),
where:
F is the force on the wire (in N),
I is the current in the wire (in A),
B is the magnetic field strength (in T),
L is the length of the wire (in m),
and theta is the angle between the current direction and the magnetic field direction.
In this case, the wire is carrying a current of 2.0 A and is at right angles to the magnetic field of 0.15 T. So, theta, the angle between the current direction and the magnetic field direction, is 90 degrees.
Substituting the given values into the formula:
F = (2.0 A) * (0.15 T) * (0.050 m) * sin(90 degrees)
Now, sin(90 degrees) is equal to 1, so we can simplify the equation:
F = (2.0 A) * (0.15 T) * (0.050 m) * 1
Multiplying the values together:
F = 0.015 N
Therefore, the magnitude of the force exerted by the magnetic field on the wire is 0.015 N.