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Chapter 19: Problem 44

A straight wire segment of length \(0.60 \mathrm{m}\) carries a current of $18.0 \mathrm{A}$ and is immersed in a uniform external magnetic ficld of magnitude \(0.20 \mathrm{T}\). (a) What is the magnitude of the maximum possible magnetic force on the wire segment? (b) Explain why the given information enables you to calculate only the maximum possible force.

Short Answer

Expert verified
Answer: The maximum magnetic force on the wire segment is 2.16 N. Only the maximum possible force can be calculated because we do not have the information about the angle between the current and the magnetic field. However, since we know the maximum value of sin胃 is 1 (when 胃 = 90掳), we can calculate the maximum possible force considering the angle to be 90掳. Any other angle would result in a lesser force, but we do not have enough information to find the exact force.

Step by step solution

01

Identify given variables

We are given the following values: - Length of the wire (L) = \(0.60 m\) - Current in the wire (I) = \(18.0 A\) - Magnitude of the uniform external magnetic field (B) = \(0.20 T\)
02

Calculate the maximum force

We can calculate the magnetic force on the wire segment using the formula: $$ F = ILB \sin\theta $$ To find the maximum value for the force, observe that \(\sin\theta\) can take values between -1 and 1, with the maximum value being \(\sin {90^\circ} =1\). So, to find the maximum force, set \(\theta = {90^\circ}\): $$ F_{max} = ILB\sin{90^\circ} = ILB $$ Now, plug in the given values and compute F_max: $$ F_{max} = (18.0\,\mathrm{A})(0.60\,\mathrm{m})(0.20\,\mathrm{T}) = 2.16\,\mathrm{N} $$ The maximum possible magnetic force on the wire segment is \(2.16\,\mathrm{N}\).
03

Explain why only the maximum possible force can be calculated

The given information does not provide the angle between the current and the magnetic field, which means the actual force could be lower than the maximum possible force, depending on the value of \(\theta\). However, since we know the maximum value of \(\sin\theta\) is 1, we can calculate the maximum possible force when \(\theta=90^\circ\). Any other angle between current and magnetic field would result in a lesser force. And thus, only the maximum possible force can be calculated with the given information.

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Most popular questions from this chapter

An early cyclotron at Cornell University was used from the 1930 s to the 1950 s to accelerate protons, which would then bombard various nuclei. The cyclotron used a large electromagnet with an iron yoke to produce a uniform magnetic field of \(1.3 \mathrm{T}\) over a region in the shape of a flat cylinder. Two hollow copper dees of inside radius \(16 \mathrm{cm}\) were located in a vacuum chamber in this region. (a) What is the frequency of oscillation necessary for the alternating voltage difference between the dees? (b) What is the kinetic energy of a proton by the time it reaches the outside of the dees? (c) What would be the equivalent voltage necessary to accelerate protons to this energy from rest in one step (say between parallel plates)? (d) If the potential difference between the dees has a magnitude of $10.0 \mathrm{kV}$ each time the protons cross the gap, what is the minimum number of revolutions each proton has to make in the cyclotron?
An electron moves with speed \(2.0 \times 10^{5} \mathrm{m} / \mathrm{s}\) in a uniform magnetic field of \(1.4 \mathrm{T},\) pointing south. At one instant, the electron experiences an upward magnetic force of $1.6 \times 10^{-14} \mathrm{N} .$ In what direction is the electron moving at that instant? Be specific: give the angle(s) with respect to $\mathrm{N}, \mathrm{S}, \mathrm{E}, \mathrm{W},$ up, down. (If there is more than one possible answer, find all the possibilitics.)
The magnetic field in a cyclotron is \(0.50 \mathrm{T}\). What must be the minimum radius of the dees if the maximum proton speed desired is $1.0 \times 10^{7} \mathrm{m} / \mathrm{s} ?$
Four long parallel wires pass through the corners of a square with side $0.10 \mathrm{m}\(. All four wires carry the same magnitude of current \)I=10.0 \mathrm{A}\( in the directions indicated. Find the magnetic field at point \)R,$ the midpoint of the left side of the square.
Two parallel wires in a horizontal plane carry currents \(I_{1}\) and \(I_{2}\) to the right. The wires each have length \(L\) and are separated by a distance \(d\). (a) What are the magnitude and direction of the ficld due to wire 1 at the location of wire \(2 ?\) (b) What are the magnitude and direction of the magnetic force on wire 2 due to this field? (c) What are the magnitude and direction of the field due to wire 2 at the location of wire \(1 ?\) (d) What are the magnitude and direction of the magnetic force on wire 1 due to this field? (e) Do parallel currents in the same direction attract or repel? (f) What about parallel currents in opposite directions?
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