Question

A DC power line for a light-rail system carries 1000 A at an angle of 30.0o to the Earth’s 5.00×10−5 -T field.

What is the force on a 100-m section of this line? (b) Discuss practical concerns this presents, if any.

What is the maximum torque on a 150-turn square loop of wire 18.0 cm on a side that carries a 50.0-A current in a 1.60-T field? (b) What is the torque when θ is 10.9o?

Answer 1

1.

Magnetic force on a current carrying wire is given by:

F = i*LxB = i*L*B*sin

Using given values:

F = 1000*100*5.00*10^-5*sin 30 deg

F = 2.5 N = Force due to magnetic field on the light-rail system

(b) Since this 2.5 N force is very low in compare to weight of light-rails system, So this is an insignificant force.

2.

Torque on a square loop in magnetic field is given by:

= N*I*A*B*sin

A = Area of loop = side^2 = 0.18^2

for max torque = = 90 deg

Using given values:

_max = 150*50.0*0.18^2*1.60*sin 90 deg

_max = 388.8 N.m

Part (b)

when = 10.9 deg, then

= 150*50.0*0.18^2*1.60*sin 10.9 deg

= 73.52 N.m

Let me know if you've any query.