/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 An equilateral triangular curren... [FREE SOLUTION] | 91Ó°ÊÓ

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An equilateral triangular current loop \(P Q R\) carries a current \(I\) ampere. Length of each side is \(l\) metre. A uniform magnetic field of induction \(\vec{B}\) exists in a direction parallel to \(P Q\). Then the force on the side \(P Q\) is (A) \(I I B\) (B) \(\frac{I l B}{2}\) (C) \(\left(\frac{I l B}{2}\right) \sqrt{3}\) (D) Zero

Short Answer

Expert verified
The magnetic force on the side \(PQ\) is zero (choice D).

Step by step solution

01

Identify the vectors involved

We have the following vectors: - \(\vec{PQ}\) is the vector from point \(P\) to point \(Q\) with a magnitude of \(l\) meters. - \(\vec{QR}\) is the vector from point \(Q\) to point \(R\) with a magnitude of \(l\) meters. - \(\vec{RP}\) is the vector from point \(R\) to point \(P\) with a magnitude of \(l\) meters. - \(\vec{B}\) is the magnetic field parallel to \(PQ\). Note that the current flows counterclockwise in the loop.
02

Find the components of \(\vec{PQ}\), \(\vec{QR}\), and \(\vec{RP}\)

Since the magnetic field is parallel to \(\vec{PQ}\), we can write the components of \(\vec{PQ}\), \(\vec{QR}\), and \(\vec{RP}\) as follows: $$\vec{PQ} = l \hat{i}$$ $$\vec{QR} = -\frac{l}{2} \hat{i} + \frac{l\sqrt{3}}{2} \hat{j}$$ $$\vec{RP} = -\frac{l}{2} \hat{i} - \frac{l\sqrt{3}}{2} \hat{j}$$
03

Calculate the magnetic force for side \(PQ\)

Using the formula for magnetic force, we can compute the magnetic force on the side \(PQ\): $$\vec{F_{PQ}} = I(\vec{PQ} \times \vec{B}) = I(l\hat{i} \times B\hat{i}) = I[0\hat{i} + 0\hat{j} + 0\hat{k}] = 0$$ The magnetic force on the side \(PQ\) is zero (choice D).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Lorentz Force Law
When we delve into the world of magnetism and electromagnetism, one fundamental principle we encounter is the Lorentz force law. This physical law describes the effect of a magnetic field on a moving electric charge. The force, which is known as the Lorentz force, is the combined effect of electric and magnetic fields on a charge. However, for a current-carrying wire, only the magnetic component is considered as the wire itself may create the electric field. The magnetic aspect of the Lorentz force is given by the equation: \[ \vec{F} = I(\vec{L} \times \vec{B}) \] where \( I \) is the current, \( \vec{L} \) is the length vector of the wire in the direction of current, and \( \vec{B} \) is the magnetic field vector. With this law, we can begin to understand how a current loop interacts with a magnetic field.
Magnetic Field
A magnetic field is a vector field that describes the magnetic influence on moving electric charges, electric currents, and magnetic materials. A magnet creates a field in the space around it that affects other magnets and objects made of magnetic materials. In our exercise, the magnetic field denoted by \( \vec{B} \) is uniform, meaning its strength and direction are the same at all points in the region. This consistency is essential when calculating the force experienced by the different sides of an equilateral triangular current loop. The field's uniformity simplifies the problem and helps us focus on how the current loop's orientation relative to the field affects the magnetic force it experiences.
Right Hand Rule
The right hand rule is a handy mnemonic that helps predict the direction of a force in a magnetic field, the direction of the magnetic field produced by a current, or the direction of the current induced in a circuit. When dealing with the magnetic force on a current-carrying wire, you point your thumb in the direction of the current and your fingers in the direction of the magnetic field; your palm then faces in the direction of the force. This rule is critical for visualizing and solving problems involving the interaction of currents and magnetic fields, like the one in our exercise. By using the right hand rule, students can better understand the three-dimensional aspects of the forces at play. Using this rule together with the Lorentz force equation can also help to ensure you've got both the direction and magnitude of the forces correct.
Equilateral Triangular Current Loop
In our example, we are dealing with an equilateral triangular current loop. This geometric shape is essential because the angle between any two sides is always 60 degrees, which plays a significant role in determining the direction of the magnetic force. When a current flows through such a loop placed in a magnetic field, forces will act on each of the three sides depending on their orientation relative to the field. Because of the equilateral triangle's symmetry, we can often make simplifications when calculating these forces. In our exercise, we see that because the side \(PQ\) is parallel to the magnetic field \(\vec{B}\), the magnetic force on side \(PQ\) is zero. This outcome, derived from the understanding of vector cross products, illustrates how the orientation of the current loop relative to the magnetic field can result in different magnitudes of force on different sides, which is central to devices like electric motors and generators.

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Most popular questions from this chapter

A long wire having linear charge density \(\lambda\) moving with constant velocity \(v\) along its length. A point charge moving with same speed in opposite direction and at that instant, it is \(r\) distance from the wire. The net force acting on the charge is given by (A) \(\frac{\lambda q}{2 \pi r}\left[\frac{1}{\varepsilon_{0}}+v^{2} \mu_{0}\right]\) (B) \(\frac{\lambda q}{2 \pi r}\left[\frac{1}{\varepsilon_{0}}-\mu_{0} v^{2}\right]\) (C) \(\frac{\lambda q}{2 \pi r} \sqrt{\left(\frac{1}{\varepsilon_{0}}\right)^{2}+v^{4} \mu_{0}^{2}}\) (D) Zero

An electron accelerated by a potential difference \(V=3.6 \mathrm{~V}\) first enters into a uniform electric field of a parallel-plate capacitor whose plates extend over a length \(l=6 \mathrm{~cm}\) in the direction of initial velocity. The electric field is normal to the direction of initial velocity and its strength varies with time as \(E=a \times t\), where \(a=\) \(3200 \mathrm{Vm}^{-1} \mathrm{~s}^{-1}\). Then the electron enters into a uniform magnetic field of induction \(B=\pi \times 10^{-9} \mathrm{~T}\). Direction of magnetic field is same as that of the electric field. Calculate pitch (in \(\mathrm{mm}\) ) of helical path traced by the electron in the magnetic field (Mass of electron, \(m=9 \times\) \(10^{-31} \mathrm{~kg}\) ). [Neglect the effect of induced magnetic field.]

A metal disc of radius \(R=6 \mathrm{~cm}\) is mounted on a frictionless axle. The current can flow through the axle out along the disc to a sliding contact of rim of the disc. A uniform magnetic field \(B=2 \mathrm{~T}\) is parallel to the axis of the disc. When the current is \(3 \mathrm{~A}\), the disc rotateswith constant angular velocity. The frictional force at the rim between the stationary electrical contact and the rotating rim is \(9 x \times 10^{-2} \mathrm{~N}\). Find the value of \(x\). W

Two identical wires \(A\) and \(B\), each of length \(\cdot \ell\) ', carry the same current I. Wire \(A\) is bent into a circle of radius \(R\) and wire \(B\) is bent to form a square of side 'a'. If \(B_{\mathrm{A}}\) and \(B_{\mathrm{B}}\) are the values of magnetic field at the centres of the circle and square respectively, then the ratio \(\frac{B_{A}}{B_{B}}\) is [2016] (A) \(\frac{\pi^{2}}{16 \sqrt{2}}\) (B) \(\frac{\pi^{2}}{16}\) (C) \(\frac{\pi^{2}}{8 \sqrt{2}}\) (D) \(\frac{\pi^{2}}{8}\)

A wire is bent to form a semi-circle of radius \(a\). The wire rotates about its one end with angular velocity \(\omega\). Axis of rotation being perpendicular to plane of the semicircle. In the space, a uniform magnetic field of induction \(\mathrm{B}\) exists along the axis of rotation as shown. The correct statement is(A) potential difference between \(\mathrm{P}\) and \(\mathrm{Q}\) is equal to \(2 B \omega a^{2}\) (B) potential difference between \(\mathrm{P}\) and \(\mathrm{Q}\) is equal to \(2 \pi^{2} B \omega a^{2}\)(C) \(P\) is at higher potential than \(Q\). (D) potential difference between \(P\) and \(Q\) is zero.

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