Question

A wire 25.0cm long lies along the z-axis and carries a current of 9.00A in the positive +z-direction. The magnetic field is uniform and has components Bx=-0.242T, By=-0.985T and Bz=-0.336T. Find the components of the magnetic force on the wire?
b) What is the magnitude of the net magnetic force on the wire?​

Answer 1

The expression of the magnetic force and solving the determinant allows to shorten the result for the value of the magnetic force are:

• In Cartesian form  F = 2.46 i ^ - 0.605 j ^

• In the form of magnitude and direction F = 2.53 N and θ = 346.2º

Given parameters.

• Length of the wire on the z axis is: L = 25.0 cm = 0.25 m.

• The current i = 9.00 A in the positive direction of the z axis.

• The magnetic field B = (-0.242 i ^ - 0.985 j ^ -0.336 k ^ ) T

To find.

• Magnetic force.

The magnetic force on a wire carrying a current is the vector product of the direction of the current and the magnetic field.

          F = i L x B

Where the bold letters indicate vectors, F is the force, i the current, L a vector pointing in the direction of the current and B the magnetic field.

The best way to find the force is to solve the determinant, in general, a vector (L) is written in the form of the module times a unit vector.

         $F= i |L| \left[\begin{array}{ccc}i&j&k\\L_x&L_y&L_z\\B_x&B_y&B_z\end{array}\right]$  

Let's calculate.

       $F= 9.00 \ 0.25 \ \left[\begin{array}{ccc}i&j&k\\0&0&1\\-0.242&-0.985&-0.336\end{array}\right]$  

       $F = 2.26 \ ( - 1 B_y i \ + 1 B_x j \ )$  

       F = 2.5 (0.985 i ^ - 0.242 j ^)

       F = ( 2.46 i ^ - 0.605 j^ ) N

To find the magnitude we use the Pythagorean theorem.

        F = $\sqrt{F_x^2 + F_y^2}$  

        F = $\sqrt{2.46^2 + 0.605^2}$  

        F = 2.53 N

Let's use trigonometry for the direction.

        Tan θ ’= $\frac{F_y}{F_x}$  

        θ'= tan⁻¹ $\frac{F_y}{F_x}$  

        θ'= tan⁻¹1 ($\frac{-.605}{2.46}$ )

        θ’= -13.8º

To measure this angle from the positive side of the x-axis counterclockwise.

        θ = 360- θ'

        θ = 360 - 13.8

        θ = 346.2º

In conclusion using the expression of the magnetic force and solving the determinant we can shorten the result for the value of the force are:

• In Cartesian form    F = 2.46 i ^ - 0.605 j ^

• In the form of magnitude and direction  F = 2.53 N and θ = 346.2º

Learn more here:  brainly.com/question/2630590