Question

A wire having a mass per unit length of 0.500 g/cm carries a2.00-A current horizontally to the south. What are the direction and magnitude of the minimum magnetic field needed to lift this wire vertically upward?

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Answer 1

(a) The current \(I\) in the wire is pointed towards the south. From the right-hand rule for currents in the magnetic field, the magnetic field \(\bar{B}\) must be towards the east to make the magnetic force \(\bar{F}_{s}\) upward. Therefore the direction of the magnetic field is towards the east.

(b) Mass per unit length of the wire is,

$$ \begin{aligned} \frac{m}{L} &=0.500 \mathrm{~g} / \mathrm{cm} \\ &=(0.500 \mathrm{~g} / \mathrm{cm})\left(\frac{10^{-3} \mathrm{~kg}}{1 \mathrm{~g}}\right)\left(\frac{1 \mathrm{~cm}}{10^{-2} \mathrm{~m}}\right) \\ &=0.05 \mathrm{~kg} / \mathrm{m} \end{aligned} $$

As the direction of the current is towards the south and the direction of the magnetic field is towards the east, the angle between them is \(\theta=90^{\circ}\). To lift the wire vertically upward, the magnetic force on the wire must be equal to its weight.

$$ \begin{aligned} \overline{F_{s}} &=\overline{F_{s}} \\ B I L \sin 90^{\circ} &=m g \\ B I L &=m g \\ B &=\left(\frac{m}{L}\right) \frac{g}{I} \end{aligned} $$

Thus, the magnitude of the magnetic field is,

$$ \begin{aligned} B &=\frac{(0.05 \mathrm{~kg} / \mathrm{m})\left(9.80 \mathrm{~m} / \mathrm{s}^{2}\right)}{2.00 \mathrm{~A}} \\ &=0.245 \mathrm{~T} \end{aligned} $$

Hence, the minimum magnetic field needed to lift the wire vertically upward is \(0.245 \mathrm{~T}\).