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Chapter 26: Problem 28

A wire with mass per unit length 75 g/m runs horizontally at right angles to a horizontal magnetic field. A 6.2 -A current in the wire results in its being suspended against gravity. What's the magnetic field strength?

Short Answer

Expert verified
The magnetic field strength is approximately \(0.119 T\).

Step by step solution

01

Convert units

Convert mass per unit length from grams per meter (g/m) to kilograms per meter (kg/m). 75 g/m converted to kg/m is 0.075 kg/m.
02

Determine the weight of the wire per unit length

The weight of the wire is obtained using the relation \(W = mg\), where \(m\) is the mass of the wire and \(g\) is the acceleration due to gravity (approximately \(9.81 m/s^2\)). The weight per unit length is therefore \(0.075 kg/m * 9.81 m/s^2 = 0.73575 N/m\).
03

Apply the formula for magnetic force.

The formula for magnetic force is \(F = BIl\), where \(F\) is the force, \(B\) is the magnetic field strength, \(I\) is the current and \(l\) is the length of the wire. But since the weight of the wire per unit length is equal to the magnetic force per unit length, we can equate \(mg = BIl\). Given that the unit lengths cancel out and knowing that the magnetic force and weight are balanced (\(mg = BI\)), we can solve for \(B\).
04

Calculate for the magnetic field strength

Rearranging the equation from the previous step to isolate \(B\), we find that \(B = mg/I\). Substituting, \(B = 0.73575 N/m / 6.2 A = 0.11867 T\).

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Most popular questions from this chapter

In what two senses does a current loop behave like a magnetic dipole?

A simplified model of Earth's magnetic field has it originating in a single current loop at the outer edge of the planet's liquid core (radius \(3000 \mathrm{km}\) ). What current would give the \(62-\mu \mathrm{T}\) field measured at the north magnetic pole?

A 2.2 -m-long wire carrying 3.5 A is wound into a tight coil \(5.0 \mathrm{cm}\) in diameter. Find the magnetic field at its center.

An unmagnetized piece of iron has no net magnetic dipole moment, yet it's attracted to either pole of a bar magnet. Why?

Your company is developing a device incorporating a \(20-\mathrm{cm}-\) diameter coil carrying 0.50 A that, when properly oriented, will just cancel Earth's \(50-\mu\) T magnetic field at the coil's center. How much wire must you requisition for each coil?
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