Question

A long iron bar lies along the x-axis and has current of I = 16.4 A running through it in the +x-direction. The bar is in the presence of a uniform magnetic field, perpendicular to the current. There is a magnetic force per unit length on the bar of 0.132 N/m in the −y-direction. (a) What is the magnitude of the magnetic field (in mT) in the region through which the current passes?

Answer 1

Answer:

B = 8.0487mT

Explanation:

To solve the exercise it is necessary to take into account the considerations of the Magnetic Force described by Faraday,

The magnetic force is given by the formula

$F =BILsin\theta$

Where,

B = Magnetic Field

I = Current

L = Length

$\theta =$ Angle between the magnetic field and the velocity, for this case are perpendicular, then is 90 degrees

According to our data we have that

I = 16.4A

F = 0.132N/m

As we know our equation must be modificated to Force per length unit, that is

$\frac{F}{L} = BI sin(90)$

Replacing the values we have that

$0.132 = 16.4 (1) B$

Solving for B,

$B = \frac{0.132}{16.4}$

$B = 8.0487mT$