/*! 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 50 A conductor in the form of a rig... [FREE SOLUTION] | 91影视

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Chapter 15: Problem 50

A conductor in the form of a right angle \(A B C\) with \(A B=3 \mathrm{~cm}\) and \(B C=4 \mathrm{~cm}\) carries a current of \(10 \mathrm{~A}\). There is a uniform magnetic field of \(5 \mathrm{~T}\) perpendicular to the plane of the conductor. The force on the conductor will be (A) \(1.5 \mathrm{~N}\) (B) \(2.0 \mathrm{~N}\) (C) \(2.5 \mathrm{~N}\) (D) \(3.5 \mathrm{~N}\)

Short Answer

Expert verified
The total magnetic force on the right-angled conductor carrying a current of 10 A and with sides AB=3 cm and BC=4 cm in a uniform magnetic field of 5 T perpendicular to the conductor is approximately \(2.5 \mathrm{~N}\).

Step by step solution

01

Calculate the magnetic force on side AB

Since we have a right-angled conductor, we can separate it into two segments and calculate the magnetic force on each segment. Let's start with side AB. To calculate the force on side AB, we use the formula F = B * I * L * sin(胃), where F is the magnetic force, B is the magnetic field, I is the current, and L is the length of the conductor. For side AB, we have: - B = 5 T (given) - I = 10 A (given) - L = 3 cm (given, but we need to convert it to meters, so L = 0.03 m) - 胃 = 90掳 (the magnetic field is perpendicular to the conductor) Now, calculating the force on side AB: F_AB = B * I * L * sin(胃) F_AB = 5 * 10 * 0.03 * sin(90掳) F_AB = 1.5 N
02

Calculate the magnetic force on side BC

Now, let's calculate the magnetic force on side BC. For side BC, we have: - B = 5 T (given) - I = 10 A (given) - L = 4 cm (given, but we need to convert it to meters, so L = 0.04 m) - 胃 = 90掳 (the magnetic field is perpendicular to the conductor) Calculating the force on side BC: F_BC = B * I * L * sin(胃) F_BC = 5 * 10 * 0.04 * sin(90掳) F_BC = 2 N
03

Combine the magnetic forces on sides AB and BC

Now we need to combine the magnetic forces on sides AB and BC to find the total magnetic force on the conductor. Since these forces are perpendicular to each other, we can use the Pythagorean theorem to find the resultant force. F_total^2 = F_AB^2 + F_BC^2 F_total^2 = (1.5 N)^2 + (2 N)^2
04

Calculate the total magnetic force on the conductor

Now we can solve for the total magnetic force on the conductor. F_total^2 = 2.25 + 4 F_total^2 = 6.25 F_total = sqrt(6.25) F_total 鈮� 2.5 N So, the total magnetic force on the conductor is approximately 2.5 N, and the correct answer is (C) \(2.5 \mathrm{~N}\).

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Most popular questions from this chapter

The magnetic field due to a current carrying circular loop of radius \(3 \mathrm{~cm}\) at a point on the axis at a distance of \(4 \mathrm{~cm}\) from the centre is \(54 \mu \mathrm{T}\). What will be its value at the centre of loop? (A) \(125 \mu \mathrm{T}\) (B) \(150 \mu \mathrm{T}\) (C) \(250 \mu \mathrm{T}\) (D) \(75 \mu \mathrm{T}\)

A coil in the shape of an equilateral triangle of side \(0.02 \mathrm{~m}\) is suspended from vertex such that it is hanging in a vertical plane between the pole-pieces of a permanent magnet producing a horizontal magnetic field of \(5 \times 10^{-2} \mathrm{~T}\). Current in loop is \(0.1 \mathrm{~A}\). Then (A) magnetic moment of the loop is \(\sqrt{3} \times 10^{-5} \mathrm{~A} / \mathrm{m}^{2}\). (B) magnetic moment of the loop is \(1 \times 10^{-6} \mathrm{~A} / \mathrm{m}^{2}\). (C) couple acting on the coil is \(5 \sqrt{3} \times 10^{-7} \mathrm{~N} / \mathrm{m}\). (D) couple acting on the coil is \(\sqrt{3} \times 10^{-7} \mathrm{~N} / \mathrm{m}\).

In which type of materials the magnetic susceptibility does not depend on temperature? (A) Diamagnetic (B) Paramagnetic (C) Ferromagnetic (D) Ferrite

Assertion: A conducting circular disc of radius \(R\) rotates about its own axis with angular velocity \(\omega\) in uniform magnetic field \(B_{0}\) along axis of the disc then no EMF is induced in the disc. Reason: Whenever a conductor cuts across magnetic lines of flux, an EMF is induced in the conductor. (A) \(\mathrm{A}\) (B) \(\mathrm{B}\) (C) \(\overline{\mathrm{C}}\) (D) D

The magnetic moment of a small current carrying loop is \(2.1 \times 10^{-25} \mathrm{~A} \times \mathrm{m}^{2}\). The magnetic field at a point on its axis at a distance of \(1 \AA\) is (A) \(4.2 \times 10^{-2} \mathrm{~Wb} / \mathrm{m}^{2}\) (B) \(4.2 \times 10^{-3} \mathrm{~Wb} / \mathrm{m}^{2}\) (C) \(4.2 \times 10^{-4} \mathrm{~Wb} / \mathrm{m}^{2}\) (D) \(4.2 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^{2}\)
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