An isolated segment of wire of length L=4.50 m carries a current of magnitude i=35.0 A at an angle theta=50.3 degrees with respect to a constant magnetic field with magnitude B= 6.70E-2 T. What is the magnitude of the magnetic force on the wire?

a) 2.66 N
b) 3.86
c) 5.60 N
d) 8.12 N
e) 11.8 N
What I did was:
F=ILxBsin(theta)
=(35.0A)(4.50m)*( 6.70E-2T * sin(50.3 degrees)
I got 8.12 N but I checked on the back of the book and it says that the answer is (e) but I don't know how to get that.

To calculate the magnitude of the magnetic force on a wire, you correctly used the formula F = ILBsinθ, where F is the magnetic force, I is the current, L is the length of the wire, B is the magnetic field strength, and θ is the angle between the direction of the current and the magnetic field.

Let's recheck your calculation step by step:

F = (35.0 A) * (4.50 m) * (6.70E-2 T * sin(50.3 degrees))

First, convert the angle from degrees to radians:
θ (in radians) = θ (in degrees) * (π / 180 degrees)
θ (in radians) = 50.3 degrees * (π / 180 degrees) ≈ 0.878 rad

Then, plug the values into the formula:
F = (35.0 A) * (4.50 m) * (6.70E-2 T * sin(0.878 rad))
F ≈ 8.12 N

Since your calculation is correct, the magnitude of the magnetic force on the wire is indeed 8.12 N, which corresponds to option (d).

If the answer in the book states that option (e) is correct, either there was an error in the book or there may be additional factors involved in the calculation that were not provided in the question.