Question

A straight wire 0.010 m long lies in the field of several horseshoe magnets. (The horseshoe magnets are lying side by side, with their magnetic fields pointing in the same direction; see diagram `Current balance and circuit' in the lab manual.) The wire is perpendicular to the magnetic field, B. The magnitude of B is uniform along the length of the wire and has the value 0.16 T. The wire carries a current, I, of 2.1 A. What is the magnitude of the force exerted on the wire by the magnetic field?

Correct, computer gets: 3.36e-03 N

2. [1pt]
For the wire and magnetic field in the previous question, the directions of I and B are shown in the table below as compass directions in a plane, for four different cases. For each case, enter the direction of the force F_B exerted on the wire by the magnetic field. E.g., if F_B is Up (out of the plane) for case 1, North for 2, Down (into the plane) for 3, and West for 4, enter UNDW.

	I	B	FB
1	N	E
2	E	N
3	N	W
4	E	S

Answer 1

a)

given, Current, I = 2.1 A

length of wire, L = 0.010 m

Magnetic field, B = 0.16 T

Magnetic force, F = IL X B = ILB sin

since wire is perpendicular to the magnetic field B, so = 90 degree

so, F = ILB

= 2.1 * 0.01 * 0.16 = 3.36 * 10^-3 N

b)
Direction of force F can be find by Right Hand Rule,

Point fingure in the direction of current I , thumb in the direction of magnetic field, then palm direction is the direction of force,

1) Up (out of the plane)

2) Down (into the plane)

3) Down (into the plane)

4)  Up (out of the plane)

so, Enter UDDU