Question

The force on a wire is a maximum of6.71 10-2 N when placed between the pole faces of a magnet.The current flows horizontally to the right and the magnetic fieldis vertical. The wire is observed to "jump" toward the observerwhen the current is turned on.(a) What type of magnetic pole is the top pole face?NorthpoleSouthpole(b) If the pole faces have a diameter of 15.0 cm, estimate the current in the wire if thefield is 0.16 T.A(c) If the wire is tipped so that it makes an angle of 10.0°with the horizontal, what force will it now feel?N

Answer 1

Answer:

B.   i=2.79A

C.   F=0.066N

Explanation:

A) By the right hand rule we have that

F=iL x B

F=iLBsin(α)

If the wire jump toward the observer the top pole face is the magnetic southpole.

B) The diameter of the pole face is 15cm. We can take this value as L (the length in which the wire perceives the magnetic field). Hence, we have

$F=iLBsin(\alpha)\\\alpha=90°\\F=iLB\\i=\frac{F}{LB}=\frac{6.71*10^{-2}N}{(0.15m)(0.16T)}=2.79A$

C) Now the length of the wire that feels B is

$L=\frac{0.15m}{cos(10\°)}=0.152m$

and the force will be (by taking the degrees between the magnetic field vector and current vector as 80°)

$F=iLBsin(\alpha)\\F=(2.79A)(0.152m)(0.16T)(sin(80\°))=0.066N$

I hope this is useful for you

regards