Question

The size of the magnetic force on a straight wire of length L carrying current I in a uniform magnetic field with strength B is

F=ILBsin(?).

Here ? is the angle between the direction of the current (along the wire) and the direction of the magnetic field. Hence Bsin(?) refers to the component of the magnetic field that is perpendicular to the wire, B?. Thus this equation can also be written as F=ILB?.

The direction of the magnetic force on the wire can be described using a "right-hand rule." This will be discussed after Part B.

Up Wイ Down

A) Consider a wire of length L = 0.30 m that runs north-south on a horizontal surface. There is a current of I = 0.50 A flowing north in the wire. (The rest of the circuit, which actually delivers this current, is not shown.) The Earth's magnetic field at this location has a magnitude of 0.50 gauss (or, in SI units, 0.5×10?4tesla) and points north and 38 degrees down from the horizontal, toward the ground. What is the size of the magnetic force on the wire due to the Earth's magnetic field? In considering the agreement of units, recall that 1T=1N/(A?m).(Figure 1)

B) Now assume that a strong, uniform magnetic field of size 0.55 T pointing straight down is applied. What is the size of the magnetic force on the wire due to this applied magnetic field? Ignore the effect of the Earth's magnetic field.

C) Now let us assume that we are able to reorientate the wire in (Figure 1) and therefore the direction of current flow. Which of the following situations would result in a magnetic force on the wire that points due north? (select all of the following that apply)

1. Current in the wire flows straight down; the magnetic field points due west.

2. Current in the wire flows straight up; the magnetic field points due east.

3.Current in the wire flows due east; the magnetic field points straight down

4.Current in the wire flows due west and slightly up; the magnetic field points due east

5.Current in the wire flows due west and slightly down; the magnetic field points straight down.

D) Assume that the applied magnetic field of size 0.55 T in (Figure 1) is rotated so that it points horizontally due south. What is the size of the magnetic force on the wire due to the applied magnetic field now?

Answer 1

Solution; Given;l Length of the wire L-0.30m I Currentflowingin the wire.I 0.50A -1 Magneticfield- B 0.50gauss or-0.5x10 teslaq Angle 38° (A)I The size of themagneticforce on the-wire-due to-earth'smagneticfield-is1 F=ILB sin θ (0.50A) (0.30m) (0.50x10 T (sin38°)1 4.6x10-5N The size-ofthe-magneticforce on-the-wire-due to-earth'smagneticfield-is 4.6x10 N.T (B)(I Nowuniform-magnetic-fieldof-size 0.55T-is-pointing-down.ף So, θ = 90 The size-of the-magneticforce on the-wire-due to-this applied-magneticfield-is F-ILB sin -(0.50A) (0.30m) (0.55T (sin 90) -4.6x10-6 N -0.083N The size-of the magneticforce on the-wire-due to this-applied-magneticfield-is 0.083N.I

Theright-handrule-helpsto-determine the directionofthe-magneticforce.1 In-right-hand rule,the thumbpoints towards the force direction, the secondfingerpoints towardsthe current direction-and the third-fingerpoints-towardsthemagnetic-fielddirection.1 The direction-of the-force,currentand.magneticfield-shouldbe-perpendicular toeach-other.1 So,the-first foursituations-wouldresultin a-magnetic-force onthe-wire that points-due north those are 1,2,3, and 4.1 The applied-magneticfield-of-0.55T is-rotated. So-the angle θ 0°-or-360° .1 The magneticforcedue to-applied.magneticfield-isT F=ILB sin θ -(0.50A) (0.30m)(0.55 T) (sin09) The size-of the-magneticforce dueto applied-magneticfield-is 0 N.T