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Chapter 27: Problem 45

A coil consists of 120 circular loops of wire of radius \(4.8 \mathrm{~cm} .\) A current of 0.49 A runs through the coil, which is oriented vertically and is free to rotate about a vertical axis (parallel to the \(z\) -axis). It experiences a uniform horizontal magnetic field in the positive \(x\) -direction. When the coil is oriented parallel to the \(x\) -axis, a force of \(1.2 \mathrm{~N}\) applied to the edge of the coil in the positive \(y\) -direction can keep it from rotating. Calculate the strength of the magnetic field.

Short Answer

Expert verified
Solution: Using the given values and calculations, we can find the strength of the magnetic field by plugging them into the equation, \(B = \frac{τ}{M}\): \(B = \frac{1.2 \cdot 0.048}{120 \cdot 0.49 \cdot \pi(0.048)^2}\) Upon solving, we get: \(B \approx 0.201 \, T\) Thus, the strength of the magnetic field acting on the circular coil is approximately \(0.201 \, T\).

Step by step solution

01

Calculate the magnetic moment (M) of the coil

The magnetic moment (M) of a coil is given by the product of the current (I), the area of the loop (A), and the number of loops (n): \(M = nIA\) We are given: - n = 120 loops - I = 0.49 A - A = πr^2, radius (r) = 4.8 cm = 0.048 m We can plug these values into the formula to find M: \(M = 120 \cdot 0.49 \cdot \pi(0.048)^2\)
02

Calculate the torque (τ) experienced by the coil

The torque experienced by a coil is given by the force (F) applied to keep the coil from rotating multiplied by the perpendicular distance (r) from the rotation axis: \(τ = Fr\) Here, we have: - F = 1.2 N - r = 0.048 m We can now calculate the torque: \(τ = 1.2 \cdot 0.048\)
03

Relate the torque (τ) to the magnetic moment (M) and the magnetic field (B)

The torque acting on a coil is also given by the product of the magnetic moment (M) and the magnetic field (B): \(τ = MB\sin\theta\) Since the coil is oriented parallel to the \(x\)-axis, the angle θ between the magnetic field and the plane of the coil is \(90^{\circ}\). The sine of \(90^{\circ}\) is 1, so we have: \(τ = MB\) Now we have enough information to solve for the magnetic field (B).
04

Calculate the magnetic field (B)

We can write the expression for B as: \(B = \frac{τ}{M}\) We can substitute the values obtained in steps 1 and 2 to find the magnetic field: \(B = \frac{1.2 \cdot 0.048}{120 \cdot 0.49 \cdot \pi(0.048)^2}\) Now, solve for B to get the strength of the magnetic field.

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Most popular questions from this chapter

A straight wire carrying a current of 3.41 A is placed at an angle of \(10.0^{\circ}\) to the horizontal between the pole tips of a magnet producing a field of \(0.220 \mathrm{~T}\) upward. The poles tips each have a \(10.0 \mathrm{~cm}\) diameter. The magnetic force causes the wire to move out of the space between the poles. What is the magnitude of that force?

A particle with a charge of \(+10.0 \mu \mathrm{C}\) is moving at \(300 \cdot \mathrm{m} / \mathrm{s}\) in the positive \(z\) -direction. a) Find the minimum magnetic field required to keep it moving in a straight line at constant speed if there is a uniform electric field of magnitude \(100 . \mathrm{V} / \mathrm{m}\) pointing in the positive \(y\) -direction. b) Find the minimum magnetic field required to keep the particle moving in a straight line at constant speed if there is a uniform electric field of magnitude \(100 . \mathrm{V} / \mathrm{m}\) pointing in the positive \(z\) -direction.

A particle with a charge of \(20.0 \mu \mathrm{C}\) moves along the \(x\) -axis with a speed of \(50.0 \mathrm{~m} / \mathrm{s}\). It enters a magnetic field given by \(\vec{B}=0.300 \hat{y}+0.700 \hat{z},\) in teslas. Determine the magnitude and the direction of the magnetic force on the particle.

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 particle with mass \(m\), charge \(q\), and velocity \(v\) enters a magnetic field of magnitude \(B\) and with direction perpendicular to the initial velocity of the particle. What is the work done by the magnetic field on the particle? How does this affect the particle's motion?
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