Question

As shown in the figure, a conducting rod with a linear mass density of 0.0395 kg/m is suspended by two flexible wires of negligible mass in a uniform magnetic field directed into the page. A power supply is used to send a current through the rod such that the tension in the support wires is zero.

(a) If the magnitude of the magnetic field is 3.70 T, determine the current in the conducting rod.

(b) Determine the direction of the current in the conducting rod.

to the left or to the right   

Answer 1

The magnetic force on a current carrying conductor in a magnetic field is given as follows:

$$ \begin{aligned} \vec{F} &=I(\vec{L} \times \vec{B}) \\ &=(B I L \sin \theta) \hat{n} \end{aligned} $$

Here, direction of \(\vec{L}\) is along the direction of current and direction of \(\hat{n}\) is direction of force vector.

(a)

Since, direction of magnetic field is perpendicular to the direction of current. The value of \(\theta\) is \(90^{\circ}\).

The magnitude of magnetic force on the wire is equal to the gravitational force on the wire.

Therefore,

\(B I L \sin \theta=m g\)

Rearrange for \(\underline{\underline{L}}\)

\(I=\frac{m g}{B L \sin \theta}\)

\(=\left(\frac{g}{B \sin \theta}\right)\left(\frac{m}{L}\right)\)

$$ =\left(\frac{9.8 \mathrm{~m} / \mathrm{s}^{2}}{(3.70 \mathrm{~T})\left(\sin 90^{\circ}\right)}\right)(0.0395 \mathrm{~kg} / \mathrm{m}) $$

\(=0.105 \mathrm{~A}\)

Therefore, current in the wire is \(0.105 \mathrm{~A}\)

(b)

The direction of the force must be upward, and it will be the direction of cross product of length vector and direction of magnetic field.

Since, the direction of magnetic field is into the page. The direction of current must be towards right.