Question

A straight, vertical wire carries a current of 2.15 A downward in a region between the poles of a large superconducting electromagnet, where the magnetic field has a magnitude of B = 0.574 T and is horizontal.

a) What is the magnitude of the magnetic force on a 1.00 cm section of the wire that is in this uniform magnetic field, if the magnetic field direction is east?

b) What is the magnitude of the magnetic force on a 1.00 cm section of the wire that is in this uniform magnetic field, if the magnetic field direction is south?

c) What is the magnitude of the magnetic force on a 1.00 cm section of the wire that is in this uniform magnetic field, if the magnetic field direction is 30.0 degrees south of west?

d) What is the direction of this magnetic force?

Answer 1

Answer:

a) F = 0.01234N to the south direction

b) F = 0.01234N to the north direction

c) F = 0.01234N

d) North of West

Explanation:

The magnetic force in a magnetic field is given by:

$F = BIlsin \theta$

Length of the wire, l = 1 cm = 0.01 m

Current flowing through the wire, I = 2.15 A

a) If the magnetic field direction is east

∅ = 90

F = 0.574 * 2.15 * 0.01

F = 0.01234 N to the south direction according to the Fleming's Right Hand Rule

b) If the magnetic field direction is south

The magnitude of the magnetic force remains the same

That is , F = BIL sin 90

F = 0.01234 N to the west

c) If the magnetic field direction is 30 degrees south

The angle between the magnetic field and the length of the wire still remains 90 degrees

Therefore the magnitude of the force still remains 0.01234

F = 0.01234 N

d) the direction of the force is the North of West

Answer 2

A) The magnitude of the magnetic force with the given parameters is; F = 1.234 × 10^(-2) N towards the south.

B) The magnitude of the magnetic force with the given parameters is; F = 1.234 × 10^(-2) N towards the east.

C) The magnitude of the magnetic force with the given parameters is; F = 1.234 × 10^(-2) N.

D) Direction of the force in C is;

30° west of North.

The formula for the magnetic force F acting on the current I that is carrying a conductor that has its length as L inside a magnetic field of Strength B is;

F = BIL sinθ

Where θ is the angle between the direction of the current and the direction of the magnetic field.

We are given;

B = 0.574 T

I = 2.15 A

A) Since L = 1 cm = 0.01 m and direction is in the east, then θ = 90° and as such;

F = 0.574 × 2.15 × 0.01 × sin 90

F = 1.234 × 10^(-2) N

From Flemming's right hand rule, the direction of the force is towards the south.

B) L = 1 cm = 0.01 m and direction is in the south, then θ = 90° and as such;

F = 0.574 × 2.15 × 0.01 × sin 90

F = 1.234 × 10^(-2) N

From Flemming's right hand rule, the direction of the force is towards the east.

C) Since L = 1 cm = 0.01 m and direction is 30° south of west but then it has no bearing on θ and as such θ remains 90°. Thus;

F = 0.574 × 2.15 × 0.01 × sin 90

F = 1.234 × 10^(-2) N

D) From Flemming's right hand rule, the direction of the force in C above is 30° west of North

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