Question

The figure below shows a section of an overhead power line that is 60.5 m long and carries a current of 2.10 kA. The current is directed to the north. The magnitude of the Earth's magnetic field at this location i 4.50 x 10-5 T. The field has a northward component and a downward component, so that it makes an angle of 65.0° with the power line. 

(a) What is the magnitude of the magnetic force (in N) on the power line? 

(b) What is the direction of the magnetic force on the power line? 

 north (in the same direction as I) 

 south (in the opposite direction as I) 

 east (out of the page)

 west (into the page)

 upward

 downward

 in the same direction as B

Answer 1

Magnetic force on a current carrying wire in magnetic field is given by:

F = I*LxB = I*L*B*sin

Given that: I = current in wire = 2.10 kA = 2100 A

L = length of wire = 60.5 m

B = magnetic field = 4.50*10^-5 T

= Angle between direction of current and magnetic field = 65.0 deg

So Using these values:

F = 2100*60.5*4.50*10^-5*sin 65.0 deg

F = 5.18 N = Magnetic force on the power line

Direction of force on current carrying wire is given by right hand rule:

According to this if we point our thumb towards the direction of current and fingers towards the direction of perpendicular component of magnetic field than direction of palm will be direction of force on the wire.

Now in given case if we point our thumb in the direction of current (meaning north) and fingers towards the direction of magnetic field (meaning downward), then our palm will be in into the page direction

So, Force on wire will be in West (into the page) direction

Correct option is D.