Respuesta :
Answer:
Magnetic force, F = 0.24 N
Explanation:
It is given that,
Current flowing in the wire, I = 4 A
Length of the wire, L = 20 cm = 0.2 m
Magnetic field, B = 0.6 T
Angle between force and the magnetic field, θ = 30°. The magnetic force is given by :
[tex]F=ILB\ sin\theta[/tex]
[tex]F=4\ A\times 0.2\ m\times 0.6\ T\ sin(30)[/tex]
F = 0.24 N
So, the force on the wire at an angle of 30° with respect to the field is 0.24 N. Hence, this is the required solution.
Explanation:
The given data is as follows.
length = 20 cm, current = 4 A
B = 0.6 T, [tex]\theta[/tex] = 30°
Hence, formula to calculate the force acting on wire is as follows.
F = [tex]IlBsin \theta[/tex]
Now, putting the given values into the above formula as follows.
F = [tex]IlBsin \theta[/tex]
= [tex]4 A \times 20 cm \times \frac{10^{-2}}{1 cm} \times 0.6 T \times Sin(30^{o})[/tex]
= 0.24 N
= 0.2 N
thus, we can conclude that force on the wire is 0.2 N.
