A horizontal power line carries a current of 5000A from south to north. Earth’s magnetic field ( 60.0µT) is directed toward the north and inclined downward at 70.0^o to the horizontal.

(a) Find the magnitude.

(b) Find the direction of the magnetic force on 100m of the line due to Earth’s field.

• Current in the conductor, i = 5000 A .
• Earth’s magnetic field, $B=60\mu T=60\times {10}^{-6}T$.
• The angle between current and field, $\theta ={70}^{°}$.
• Length of the conductor, L = 100m .

By using the formula of magnetic force in terms of current and magnetic field and substituting the given quantities, we can find the magnetic force. We can find its direction by using the right-hand rule.

Formula: Magnetic force, $F_B=iLB\sin \theta$

Magnetic force in vector form,$\stackrel{\rightharpoonup }{F_B}=i\stackrel{\rightharpoonup }{L}\times \stackrel{\rightharpoonup }{B}$

The magnetic forceis given by:

$F_B=iLB\sin \theta$

Where I = Current, L = Length of the conductor, B = Magnetic field, θ = Angle between current and field.

Therefore,

$\begin{array}{rcl}{F}_B& =& \left(5000\right)\times \left(100\right)\times \left(60.0\times {10}^{-6}\right)\times \sin {70}^{\circ }\\ & =& 28.2N\end{array}$

The magnitude of magnetic force is 28.2 N.

Magnetic force in vector form is given as:

$\stackrel{\rightharpoonup }{F_B}=i\stackrel{\rightharpoonup }{L}\times \stackrel{\rightharpoonup }{B}$

We can apply the right-hand rule to this vector product.

Since the current is towards the north, the force will be towards the west direction.