A 0.0150-m wire oriented horizontally between the poles of an electromagnet carries a direct current of 9.5 A. The angle between the direction of the current and that of the magnetic field is 25.0 degrees. If the magnetic field strength is 0.845 T, what is the magnitude and direction of the magnetic force on the wire between the poles?

Magnetic force (F) exerted on a current carrying wire of length (L) and carrying current (I) in a magnetic field (B) with angle θ between the direction of current and magnetic field is given by

F = (B)(I)(L) sin θ

F = ?

B = 0.845 T

I = 9.5 A

L = 0.0150 m

θ = 25°

F = (0.845)(9.5)(0.0150) sin 25°

F = 0.0509 N

The direction of this force is perpendicular to the plane containing the magnetic field and the direction of the current.

Hope this Helps!!!

Lanuel Lanuel · Answer 2 · 2022-01-20T10:51:09+01:00

The magnitude and direction of the magnetic force on the wire between the poles is 0.051 Newton.

Given the following data:

Length of wire = 0.0150 m

Current = 9.5 A.

Angle = 25.0 degrees.

Magnetic field strength = 0.845 T

To calculate the magnitude and direction of the magnetic force on the wire between the poles:

The force in a magnetic field.

Mathematically, the magnetic force acting on a current in a magnetic field is given by this formula:

[tex]F = BILsin\theta[/tex]

Where:

B is the magnetic field strength.

I is the current flowing through a conductor.

L is the length of conductor.

[tex]\theta[/tex] is the angle between a conductor and the magnetic field.

Substituting the given parameters into the formula, we have;

[tex]F = 0.845 \times 9.5 \times 0.0150 \times sin25\\\\F = 0.1204125 \times 0.4226[/tex]

F = 0.051 Newton.

Note: The direction of the magnetic force is perpendicular to the magnetic field plane and current.

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