Current in a Wire The force on a wire is a maximum of 5.92 N when placed between

Question

Current in a Wire The force on a wire is a maximum of 5.92 N when placed between the pole faces of a magnet. The current flows horizontally to the right and the magnetic field is vertical. The wire is observed to "jump" toward the observer when the current is turned on. If the pole faces have a diameter of 11.5 cm, estimate the current in the wire if the field is 0.106 T. ????! Tries 0/30 If the wire is tipped so that it now makes an angle of 17.9° with the horizontal, what force will it now feel? ????! Tries 0/30

Explanation / Answer

Part A

Magnetic force is given by:

Fmag = i*LxB

Fmag = i*L*B*sin theta

i = Fmag/(L*B*sin theta)

theta = Angle between wire and magnetic field = 90 deg

Using givne values:

i = 5.92/(11.5*10^-2*0.106*sin 90 deg)

i = 485.64 Amp

Part B

Now theta = 90 - 17.9 deg = 72.1 deg

We need angle between wire and vertical magnetic field, given angle is between wire and horizontal axis, So we get theta by subtracting it from 90 deg

F = i*L*B*sin theta

F = 485.64*11.5*10^-2*0.106*sin 72.1 deg

F = 5.63 N

Please Upvote.

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.