Question

A straight wire is aligned east-west in a region where Earth's magnetic field has magnitude \(0.48 \mathrm{mT}\) and direction \(72^{\circ}\) below the horizontal, with the horizontal component directed due north. The wire carries a current \(I\) toward the west. The magnetic force on the wire per unit length of wire has magnitude \(0.020 \mathrm{N} / \mathrm{m}\) (a) What is the direction of the magnetic force on the wire? (b) What is the current \(I ?\)

Short answer

Solution: (a) The direction of the magnetic force on the wire has components upward and toward due north. (b) The current flowing through the wire is 41.6 A.

Step-by-step solution

Apply the right-hand rule to determine the direction of the magnetic force

We can use the right-hand rule to determine the direction of the magnetic force on the wire. Point the thumb of your right hand in the direction of the current (west in this case). Curl your fingers in the direction of the magnetic field. The direction your palm is pointing is the direction of the magnetic force. In this case, the magnetic field has components both downward and toward due north, so the magnetic force will have components upward and toward due north.

Find the magnitude of the horizontal component of the magnetic force

We are given the magnitude of the Earth's magnetic field (\(B = 0.48\ \mathrm{mT}\)) and its angle \(72^\circ\) below the horizontal. We can find the horizontal component (\(B_h\)) using the formula \(B_h = B \cos\theta,\) where \(\theta = 72^\circ\). So, \(B_h = 0.48\ \mathrm{mT} \cos 72^\circ = 0.48\ \mathrm{mT} \times 0.3090 = 0.1483\ \mathrm{mT}\).

Find the magnitude of the vertical component of the magnetic force

We can find the vertical component (\(B_v\)) of the magnetic field using the formula \(B_v = B \sin\theta,\) where \(\theta = 72^\circ\). So, \(B_v = 0.48\ \mathrm{mT} \sin 72^\circ = 0.48\ \mathrm{mT} \times 0.9511 = 0.4565\ \mathrm{mT}\).

Calculate the current in the wire

We can use the magnetic force equation for a current-carrying wire to find the current \(I\): \(F = BIL,\) where \(F\) is the force per unit length, \(B\) is the magnitude of the magnetic field, \(L\) is the length of the wire, and \(I\) is the current. In this case, the force per unit length is given as \(F = 0.020\ \mathrm{N/m}\). The force due to the horizontal component of the magnetic field is equal to \(F_h = B_hIL \Rightarrow I = \frac{F_h}{B_hL}\). The force due to the upward (vertical) component of the magnetic field is equal to \(F_v = B_vIL \Rightarrow I = \frac{F_v}{B_vL}\). Balancing these two forces, we have: \[ \frac{F_h}{B_hL} = \frac{F_v}{B_vL} \Rightarrow \frac{F_h}{B_h} = \frac{F_v}{B_v} \Rightarrow F_h = F \times \frac{B_h}{B}. \] We can plug in the values obtained in steps 2 and 3 to calculate the force due to the horizontal component of the magnetic field: \[ F_h = 0.020\ \mathrm{N/m} \times \frac{0.1483\ \mathrm{mT}}{0.480\ \mathrm{mT}} = 0.00617\ \mathrm{N/m}. \] Now, we can calculate the current in the wire using the equation for magnetic force on a current-carrying wire and the force due to the horizontal component of the magnetic field: \[ I = \frac{F_h}{B_hL} = \frac{0.00617\ \mathrm{N/m}}{0.1483\ \mathrm{mT} \times 1\ \mathrm{m}} = 41.6\ \mathrm{A}. \]

Present the answers

(a) The direction of the magnetic force on the wire has components upward and toward due north. (b) The current flowing through the wire is \(I = 41.6\ \mathrm{A}.\)