1.33 x 10⁻ ⁴ T outwards.
The equation for the magnetic force (F) on a wire whose length is L and carrying a current I in a magnetic field (B) that is uniform is given by;
F = ILB sin θ ---------------------(i)
Where;
θ = angle between the direction of the current and that of the magnetic field.
From the question,
F = 4.0 × 10⁻² N
I = 12A
L = 25m
θ = 90°
Substitute these values into equation(i) and solve as follows;
4.0 × 10⁻² = 12 x 25 x B x sin 90°
4.0 × 10⁻² = 300 x B x 1
4.0 × 10⁻² = 300B
0.04 = 300B
B = [tex]\frac{0.04}{300}[/tex]
B = 0.000133
B = 1.33 x 10⁻ ⁴ T
To get the direction of the magnetic field, the right-hand rule is used.
If the right hand fingers are positioned in the correct order specified by the right hand rule, then it would be seen that the magnetic field is directed outwards.
Therefore, the magnitude and direction of the magnetic field at this location is 1.33 x 10⁻ ⁴ T outwards.
