a current-carrying wire is oriented at right angles to a uniform magnetic field. if the lenght of the wire is 0.50m and it experiences a force at 0.90n while carrying a current of 4.0a. what is the strength of the magnetic field?

The answer is .45 T

F = B(I)L
B = F/(I)L
B = .90/4(.5)
B = .45 T
(T is Teslas)

To find the strength of the magnetic field, we can use the formula for the magnetic force experienced by a current-carrying wire in a magnetic field:

F = BIL

Where:
F is the force (0.90 N)
B is the magnetic field strength (what we're trying to find)
I is the current (4.0 A)
L is the length of the wire (0.50 m)

Rearranging the formula, we can solve for B:

B = F / (IL)

Substituting the given values:

B = 0.90 N / (4.0 A * 0.50 m)

Now we can calculate the magnetic field strength:

B = 0.90 N / 2.0 A * 0.50 m

B = 0.90 N / 1.0 A * 0.50 m

B = 0.90 N / 0.50 A * 0.50 m

B = (0.90 N * 2.0 m) / (0.50 A)

B = 1.80 Nm / 0.50 A

B = 3.60 T

Therefore, the strength of the magnetic field is 3.60 Tesla.