Question

What is the magnetic force on a 200 cm length of (straight) wire carrying a current of 30 A in a region where a uniform magnetic field has a magnitude of 0.055 T and is directed at an angle of 20 away from the wire

Answer 1

Answer:

F = 0.112 N

Explanation:

To find the magnitude of magnetic force on the wire, you use the following formula:

$|\vec{F}|=|i\vec{L}\ X\ \vec{B}|=iLBsin\theta$   (1)

L: length of the wire = 200cm = 0.2m

i: current in the wire = 30 A

B: magnitude of the magnetic field = 0.055 T

θ: angle between the directions of L and B = 20°

You replace the values of L, i, B and θ in the equation (1):

$|\vec{F}|=(30A)(0.2m)(0.055T)sin(20\°)=0.112N$

hence, the magnetic force on teh wire is 0.112N