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Chapter 27: Problem 36

A straight wire of length \(2.00 \mathrm{~m}\) carries a current of \(24.0 \mathrm{~A} .\) It is placed on a horizontal tabletop in a uniform horizontal magnetic field. The wire makes an angle of \(30.0^{\circ}\) with the magnetic field lines. If the magnitude of the force on the wire is \(0.500 \mathrm{~N}\), what is the magnitude of the magnetic field?

Short Answer

Expert verified
Answer: The magnitude of the magnetic field is approximately 0.0208 T (tesla).

Step by step solution

01

Write down the known values

We are given the following values: - Length of the wire (\(L\)) = 2.00 m - Current in the wire (\(I\)) = 24.0 A - Angle between the wire and magnetic field lines (\(\theta\)) = 30.0掳 - Force on the wire (\(F\)) = 0.500 N
02

Convert the angle to radians

In order to use the formula, we need to convert the angle from degrees to radians. We can do this using the following conversion: 1 radian = 180掳/蟺. So, we have: $$ \theta_{rad} = 30.0掳 \cdot \frac{\pi}{180掳} = \frac{\pi}{6} \, \text{radians} $$
03

Use the formula

Now that we have all the necessary values, we can use the formula for the force on a wire in a magnetic field, \(F = BIL\sin{\theta}\), to find the magnitude of the magnetic field, \(B\): $$ B = \frac{F}{IL\sin{\theta}} $$
04

Substitute the known values

Now, substitute the known values into the formula: $$ B = \frac{0.500 N}{(24.0 A)(2.00 m) \sin(\frac{\pi}{6})} $$
05

Calculate the magnetic field magnitude

Finally, calculate the magnetic field magnitude: $$ B = \frac{0.500 N}{(24.0 A)(2.00 m) \, (0.5)} = \frac{0.500 N}{24.0 A} $$ The magnitude of the magnetic field is approximately: $$ B \approx 0.0208 \, \text{T} $$ So the magnitude of the magnetic field is approximately 0.0208 T (tesla).

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Most popular questions from this chapter

An electron moves in a circular trajectory with radius \(r_{\mathrm{i}}\) in a constant magnetic field. What is the final radius of the trajectory when the magnetic field is doubled? a) \(\frac{r_{i}}{4}\) b) \(\frac{r_{i}}{2}\) c) \(r_{i}\) d) \(2 r_{i}\) e) \(4 r_{\mathrm{i}}\)

A small aluminum ball with a mass of \(5.00 \mathrm{~g}\) and a charge of \(15.0 \mathrm{C}\) is moving northward at \(3000 \mathrm{~m} / \mathrm{s}\). You want the ball to travel in a horizontal circle with a radius of \(2.00 \mathrm{~m}\), in a clockwise sense when viewed from above. Ignoring gravity, what is the magnitude and the direction of the magnetic field that must be applied to the aluminum ball to cause it to have this motion?

The work done by the magnetic field on a charged particle in motion in a cyclotron is zero. How, then, can a cyclotron be used as a particle accelerator, and what essential feature of the particle's motion makes it possible? A blue problem number indicates a worked-out solution is available in the Student Solutions Manual. One \(\bullet\) and two \(\bullet\) indicate increasing level of problem difficulty.

A straight wire with a constant current running through it is in Earth's magnetic field, at a location where the magnitude is \(0.43 \mathrm{G}\). What is the minimum current that must flow through the wire for a 10.0 -cm length of it to experience a force of \(1.0 \mathrm{~N} ?\)

A proton moving at speed \(v=1.00 \cdot 10^{6} \mathrm{~m} / \mathrm{s}\) enters a region in space where a magnetic field given by \(\vec{B}=\) \((-0.500 \mathrm{~T}) \hat{z}\) exists. The velocity vector of the proton is at an angle \(\theta=60.0^{\circ}\) with respect to the positive \(z\) -axis. a) Analyze the motion of the proton and describe its trajectory (in qualitative terms only). b) Calculate the radius, \(r\), of the trajectory projected onto a plane perpendicular to the magnetic field (in the \(x y\) -plane). c) Calculate the period, \(T,\) and frequency, \(f\), of the motion in that plane. d) Calculate the pitch of the motion (the distance traveled by the proton in the direction of the magnetic field in 1 period).
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