Question

A 6.0 m wire carries a current of 7.0 a toward the +x direction. a mag- netic force of 7.0 × 10−6 n acts on the wire in the −y direction. find the magnitude and direction of the magnetic field producing the force.

Answer 1

1) Magnitude

The magnetic force acting on a current-carrying wire is

`F=ILB \sin \theta`
where
I is the current
L is the length of the wire
B is the intensity of the magnetic field

`\theta` is the angle between the direction of the current and the direction of the magnetic field.

In our problem, the current and the force are perpendicular to each other, this means that B is perpendicular to I as well, so this angle is
`90^(\circ)` and so
`\sin \theta=1`, so we can ignore it in the formula and rewrite it as

`F=ILB`
which can be rewritten as

`B= (F)/(IL)`
and by using the data of the problem, we find the intensity of the magnetic field:

`B= (7 \cdot 10^(-6)N)/((7 A)(6.0 m))= 1.7 \cdot 10^(-7) T`

2) Direction

we can find the direction of the magnetic field by using the right-hand rule. Let's assign one vector to each finger:
index finger --> I
middle finger --> B
thumb --> F
if we take the index finger (I) in the x+ direction and the thumb (F) in the y- direction, we find that the middle finger (B) is in the z+ direction, so this is the direction of the magnetic field.