Question

A stiff wire 44.0 cm long is bent at a right angle in the middle. One section lies along the z axis and the other is along the line y=2x in the xy plane. A current of 20.5 A flows in the wire-down the z axis and out the line in the xy plane. The wire passes through a uniform magnetic field given by B =(0.318i)T.
Required:
Determine the magnitude and the direction of the total force on the wire.

Answer 1

The magnitude of force acting on wire will be 1.90 N and the direction of the force acting on the wire will be 41.9 degrees below the negative y axis.

Explanation:

It is known that the force acting on a current carrying conductor placed in a magnetic field is

$F=BIL sin theta$

Here B is the magnetic field, I is the current flowing through the wire and L is the length of the wire which is given as 44 cm.

Since the wire is bended in the middle at right angle so the length of the two sides of the wire will be 22 cm each. Also one part is lying over z axis and another part lies in the plane of xy in the equation of line y = 2x. So the slope of this wire will be

$\frac{y}{x} =2$

This will be equal to tan θ.

So θ = tan⁻¹ (2) =63.4°

Then, the length of the wire will be written as components of i, j and k.

$L = (-22)k+(22) cos ( 63.4) i+(22) sin (63.4)j$

$L = 0.098 i+0.197 j-0.22k$

Then,

F = I (L × B)

$F = 20.5 ((0.0985 i + 0.197 j -0.22k) * (0.316 i))$

$F = 20.5 (\left[\begin{array}{ccc}i&j&k\\0.098&0.197&-0.22\\0.316&0&0\end{array}\right] )$

$F = 20.5(i(0)-j(0-(-0.22*0.316))+k(0-(0.316*0.197))) = 20.5(-0.069 j-0.062 k)$

$F = -1.415 j-1.271 k$

The magnitude of force on the wire will be

$F = \sqrt{(-1.415)^{2}+(-1.27)^{2} } = \sqrt{3.615}=1.90 N$

And the direction can be found by the tan inverse of the ratio of k component to j component of the force.

$theta = tan-1(\frac{-1.271}{-1.415})= 41.9 degrees$

So the magnitude of force acting on wire will be 1.90 N and the direction of the force acting on the wire will be 41.9 degrees below the negative y axis.