/*! 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 43 A current of \(5 \mathrm{~A}\) i... [FREE SOLUTION] | 91影视

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Chapter 15: Problem 43

A current of \(5 \mathrm{~A}\) is passed through a straight wire of length \(6 \mathrm{~cm}\); then the magnetic induction at a point \(5 \mathrm{~cm}\) from the both end of the wire is (A) \(0.25\) gauss (B) \(0.125\) gauss (C) \(0.15\) gauss (D) \(0.30\) gauss

Short Answer

Expert verified
The magnetic induction at the given point is 0.4 Gauss.

Step by step solution

01

Identify given values

The current (\(I\)) is 5 Ampere, length of wire is 6 cm and the point where we need to calculate field is 5 cm away from both ends. Convert everything into SI units for ease of calculations. Thus, \(I = 5A\), \(r = 0.05m\), and length of wire doesn't matter in this context.
02

Use Biot-Savart law to calculate magnetic field at the point due to one part of the wire

The magnetic field (B) at a point due to current in a very small length of wire can be calculated using the formula \(B=\frac{\mu I}{2 \pi r}\), where \(\mu = 4\pi * 10^{-7} Tm/A\) is the permeability of free space. So substituting values, we get \(B = \frac{4\pi * 10^{-7} * 5}{2\pi * 0.05}\). Calculating it gives \(B = 2 * 10^{-5} Tesla = 0.2 Gauss\)
03

Calculate total magnetic field at the point

The above-calculated field is due to current in a small section of the wire from one end. Since the point is equidistant from both ends, contribution of field from the other end will be same. Thus, the total magnetic field will be \(B_{total} = B_{part1} + B_{part2} = 0.2 Gauss + 0.2 Gauss = 0.4 Gauss\)

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Most popular questions from this chapter

A conductor \(A B\) carries a current \(i\) in a magnetic field \(\vec{B}\). If \(\overrightarrow{A B}=\vec{r}\) and the force on the conductor is \(\vec{F}\) (A) \(\vec{F}\) does not depend on the shape of \(A B\). (B) \(\vec{F}=i(\vec{r} \times \vec{B})\). (C) \(\vec{F}=i(\vec{B} \times \vec{r})\). (D) \(|\vec{F}|=i(\vec{r} \cdot \vec{B})\).

Velocity and acceleration vector of a charged particle moving in a magnetic field at some instant are \(\vec{v}=3 \hat{i}+4 \hat{j}\) and \(\vec{a}=2 \hat{i}+x \hat{j}\). Select the correct alternative (s) (A) \(x=-1.5\) (B) \(x=3\) (C) Magnetic field is along \(z\)-direction (D) Kinetic energy of the particle is constant

Two long parallel wires \(P\) and \(Q\) are held perpendicular to the plane of the paper at a separation of \(5 \mathrm{~m}\). If \(P\) and \(Q\) carry currents of \(2.5 \mathrm{~A}\) and \(5 \mathrm{~A}\), respectively, in the same direction, then the magnetic field at a point midway between \(P\) and \(Q\) is (A) \(\frac{\mu_{0}}{\pi}\) (B) \(\frac{\sqrt{3} \mu_{0}}{\pi}\) (C) \(\frac{\mu_{0}}{2 \pi}\) (d) \(\frac{3 \mu_{0}}{2 \pi}\)

A conducting \(\operatorname{rod} A B\) of length \(l=1 \mathrm{~m}\) is moving at a velocity \(v=4 \mathrm{~m} / \mathrm{s}\) making an angle \(30^{\circ}\) with its length. \(C\) is the middle point of the rod. A uniform magnetic field \(B=2 \mathrm{~T}\) exists in a direction perpendicular to the plane of motion, then (A) \(v_{A}-v_{B}=8 \mathrm{v}\) (B) \(v_{A}-v_{B}=4 \mathrm{v}\) (C) \(v_{B}-v_{C}=2 \mathrm{v}\) (D) \(v_{B}-v_{C}=-2 \mathrm{v}\)

Three particles each of mass \(m\) and charge \(q\) are attached to the vertices of a triangular frame, made up of three light rigid rods of equal length \(l\). The frame is rotated at constant angular speed \(\omega\) about an axis perpendicular to the plane of the triangle and passing through its centre. The ratio of the magnetic moment of the system and its angular momentum about the axis of rotation is (A) \(\frac{q}{2 m}\) (B) \(\frac{q}{m}\) (C) \(\frac{2 q}{m}\) (D) \(\frac{4 q}{m}\)
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