A straight current-carrying wire is placed horizontally on an east-west line in a region of uniform magnetic field. The magnetic field here is due north, parallel to the ground, with a strength of 0.47 T. The wire has a mass per unit length of 1 g/cm.

How much current is needed to be passed through the wire to keep the wire afloat.

Write your answer in the units of A in decimal form with two digits to the right of the decimal point (e.g. 5.32); do not write any units or sign.

Question

A straight current-carrying wire is placed horizontally on an east-west line in a region of uniform magnetic field. The magnetic field here is due north, parallel to the ground, with a strength of 0.47 T. The wire has a mass per unit length of 1 g/cm.

How much current is needed to be passed through the wire to keep the wire afloat.

Write your answer in the units of A in decimal form with two digits to the right of the decimal point (e.g. 5.32); do not write any units or sign.

BlackTom AI · Accepted Answer

We start by understanding the balance of forces on the wire. To float, the magnetic force acting upward on the current-carrying wire must balance its weight per unit length.

- Mass per unit length is given as 1 g/cm. Converting to SI units: 1 g/cm = 100 g/m = 0.1 kg/m. The weight per unit length is then (mass per length) × g = 0.1 kg/m × 9.8 m/s^2 = 0.98 N/m.
- The magnetic force per unit length on a straight wire in a uniform magnetic field is F/B = I B for perpendicular orientation, where I is the current and B is the magnetic field strength. The magnitude of the force per length is I × B when the wire, magnetic field, and current are at right angles in the appropriate configuration.
- Here the wire lies horizontally, the magnetic field is directed north and horizontal, and the current runs along east–west. The magnetic force on the wire will act vertically (upward or downward) depending on the direction of I, and we assume the direction that provides an upward force to balance gravity.

To achieve equilibrium (F_m per length = weight per length): I × B = weight per length.

Plugging in B = 0.47 T and weight per length = 0.98 N/m:
I = 0.98 N/m ÷ 0.47 T ≈ 2.085106... A.

Rounding to two decimal places as requested: I ≈ 2.09 A.

Therefore, the necessary current is approximately 2.09 A. If any step seems off, recheck unit conversions: mass per length from 1 g/cm to 0.1 kg/m, weight per length 0.98 N/m, and use I = (weight per length)/B with B = 0.47 T to get the result. The key idea is balancing magnetic lift, F_m = I B, against the weight, F_g = (mass per length) × g per unit length.

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