/*! 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 55 A triangular loop of side \(l\) ... [FREE SOLUTION] | 91影视

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Chapter 15: Problem 55

A triangular loop of side \(l\) carries a current \(I .\) It is placed in a magnetic field \(B\) such that the plane of the loop is in the direction of \(B\). The torque on the loop is (A) Zero (B) \(I B l^{2}\) (C) \(\frac{\sqrt{3}}{2} I B l^{2}\) (D) \(\frac{\sqrt{3}}{4} I B l^{2}\)

Short Answer

Expert verified
The torque experienced by the triangular loop is zero. Therefore, the correct answer is (A) Zero.

Step by step solution

01

Determine \(\theta\) between the magnetic field and the plane of the loop

According to the problem statement, the magnetic field direction and the loop plane are in the same direction. That means the angle between them, \(\theta\), is equal to \(0^{\circ}\). #Step 2: Calculate the area of the triangular loop#
02

Determine the area \(A\) of the triangular loop

The triangular loop has side length \(l\). For an equilateral triangle, we can calculate the area using the formula: \[A = \frac{\sqrt{3}}{4}l^2\]. #Step 3: Calculate the torque experienced by the loop #
03

Apply the torque formula

Now that we have the angle \(\theta = 0^{\circ}\) and the area \(A = \frac{\sqrt{3}}{4}l^2\), we can use the formula for torque, \(\tau = nIA\sin{\theta}\). Since we have only one triangular loop, \(n = 1\). Plug in the values and simplify: \[\tau = (1)(I)\left(\frac{\sqrt{3}}{4} l^{2}\right)\sin{(0^{\circ})}\] Since \(\sin{0^{\circ}} = 0\): \[\tau = 0\] The torque experienced by the triangular loop is zero. Therefore, the correct answer is (A) Zero.

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Most popular questions from this chapter

A proton of mass \(1.67 \times 10^{-27} \mathrm{~kg}\) and charge \(1.6 \times\) \(10^{-19} \mathrm{C}\) is projected with a speed of \(2 \times 10^{6} \mathrm{~m} / \mathrm{s}\) at an angle of \(60^{\circ}\) to the \(x\)-axis. If a uniform magnetic field of \(0.104 \mathrm{~T}\) is applied along \(y\)-axis, the path of proton is (A) a circle of radius \(=0.2 \mathrm{~m}\) and time period \(\pi \times 10^{-7} \mathrm{~s}\). (B) a circle of radius \(=0.1 \mathrm{~m}\) and time period \(2 \pi \times 10^{-7} \mathrm{~s}\). (C) a helix of radius \(=0.1 \mathrm{~m}\) and time period \(2 \pi \times 10^{-7} \mathrm{~s}\). (D) a helix of radius \(=0.2 \mathrm{~m}\) and time period \(4 \pi \times 10^{-7} \mathrm{~s}\).

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