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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° - 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°/π. So, we have: $$ \theta_{rad} = 30.0° \cdot \frac{\pi}{180°} = \frac{\pi}{6} \, \text{radians} $$
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 is moving at \(v=6.00 \cdot 10^{7} \mathrm{~m} / \mathrm{s}\) perpendicular to the Earth's magnetic field. If the field strength is \(0.500 \cdot 10^{-4} \mathrm{~T}\), what is the radius of the electron's circular path?

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).

A copper wire with density \(\rho=8960 \mathrm{~kg} / \mathrm{m}^{3}\) is formed into a circular loop of radius \(50.0 \mathrm{~cm} .\) The cross-sectional area of the wire is \(1.00 \cdot 10^{-5} \mathrm{~m}^{2},\) and a potential difference of \(0.012 \mathrm{~V}\) is applied to the wire. What is the maximum angular acceleration of the loop when it is placed in a magnetic field of magnitude \(0.25 \mathrm{~T}\) ? The loop rotates about an axis through a diameter.

A simple galvanometer is made from a coil that consists of \(N\) loops of wire of area \(A .\) The coil is attached to a mass, \(M\), by a light rigid rod of length \(L\). With no current in the coil, the mass hangs straight down, and the coil lies in a horizontal plane. The coil is in a uniform magnetic field of magnitude \(B\) that is oriented horizontally. Calculate the angle from the vertical of the rigid rod as a function of the current, \(i\), in the coil.

An electron in a magnetic field moves counterclockwise on a circle in the \(x y\) -plane, with a cyclotron frequency of \(\omega=1.2 \cdot 10^{12} \mathrm{~Hz}\). What is the magnetic field, \(\vec{B}\) ?
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