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Chapter 13: Problem 1861

An element \(\mathrm{d} \ell^{-}=\mathrm{dx} \uparrow\) (where $\mathrm{dx}=1 \mathrm{~cm}$ ) is placed at the origin and carries a large current \(\mathrm{I}=10 \mathrm{Amp}\). What is the mag. field on the Y-axis at a distance of \(0.5\) meter ? (a) \(2 \times 10^{-8} \mathrm{k} \wedge \mathrm{T}\) (b) \(4 \times 10^{8} \mathrm{k} \wedge \mathrm{T}\) (c) \(-2 \times 10^{-8} \mathrm{k} \wedge \mathrm{T}\) (d) \(-4 \times 10^{-8} \mathrm{k} \wedge \mathrm{T}\)

Short Answer

Expert verified
The magnetic field on the Y-axis at a distance of 0.5 meters is \(-2 \times 10^{-8} \mathrm{k} \wedge \mathrm{T}\).

Step by step solution

01

Biot-Savart Law

The Biot-Savart Law states that the magnetic field (dB) created by a small element \(d\ell\), carrying a current I, can be found by the following formula: \[dB= \frac{\mu_0}{4\pi}\cdot \frac{I \cdot d\ell \times \hat{r}}{r^2}\] where \(\mu_0\) is the permeability of free space (\(4\pi \times 10^{-7} Tm/A\)), \(d\ell\) is the length of the element, \(\hat{r}\) is the unit vector pointing from the element to the point of interest, and r is the distance from the element to the point of interest.
02

Set up the Biot-Savart Integral

Since we have a small element \(d\ell^-\) at the origin, the integral for finding the magnetic field at a point on the Y-axis can be set up as: \[B = \int dB = \int \frac{\mu_0}{4\pi}\cdot \frac{I \cdot d\ell \times \hat{r}}{r^2}\]
03

Calculate the vector cross product

Since \(d\ell\)= dx⇑ and the point P is on the positive Y-axis, the unit vector \(\hat{r}\) is in the positive Y direction as well. Evaluating the cross product dℓ × \(\hat{r}\) gives: \(d\ell \times \hat{r}\) = (dx⇑) × (0,1,0) = (0,0,-dx)
04

Determine the distance

Since the point P is on the positive Y-axis at a distance of 0.5 meters (50 cm) from the origin: \(r = 50\ cm\)
05

Evaluate the Biot-Savart integral

Now, we plug the values of the cross product and distance back into the integral: \[B = \int dB = \int \frac{\mu_0}{4\pi}\cdot \frac{I \cdot (0,0,-dx)}{50^2}\] \[B_z = -\int_{0}^{1} \frac{\mu_0 \cdot I \cdot dx}{4\pi \cdot 50^2}\] Finishing the integral, we get: \[B_z = - \frac{\mu_0 I}{4\pi \cdot 50^2}\] Now, plug in the values for \(\mu_0\) and I: \[B_z = - \frac{4\pi \times 10^{-7} \: Tm/A \cdot 10 \: A}{4\pi \cdot 50^2}\]
06

Solve for the magnetic field

Simplifying and solving for the magnetic field, we get: \[B_z = -2 \times 10^{-8} \: T\] Therefore, the correct answer is (c) \(-2 \times 10^{-8} \mathrm{k} \wedge \mathrm{T}\).

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Most popular questions from this chapter

A long straight wire carrying current of \(30 \mathrm{Amp}\) is placed in an external uniform mag. field of induction \(4 \times 10^{-4}\) tesla. The mag. field is acting parallel to the direction of current. The magnitude of the resultant magnetic induction in tesla at a point \(2 \mathrm{~cm}\) away from the wire is tesla. (a) \(10^{-4}\) (c) \(5 \times 10^{-4}\) (b) \(3 \times 10^{-4}\) (d) \(6 \times 10^{-4}\)

In a H-atom, an electron moves in a circular orbit of radius $5.2 \times 10^{-11}\( meter and produces a mag. field of \)12.56$ Tesla at its nucleus. The current produced by the motion of the electron will be (a) \(6.53 \times 10^{-3}\) (b) \(13.25 \times 10^{-10}\) (c) \(9.6 \times 10^{6}\) (d) \(1.04 \times 10^{-3}\)

In each of the following questions, Match column-I and column-II and select the correct match out of the four given choices.(B) Ammeter (Q) Moderate resistance (C) Voltmeter (R) High, Low or moderate resistance (D) Avometer (S) High resistance (a) $\mathrm{A} \rightarrow \mathrm{P} ; \mathrm{B} \rightarrow \mathrm{Q} ; \mathrm{C} \rightarrow \mathrm{R} ; \mathrm{D} \rightarrow \mathrm{S}$ (b) $\mathrm{A} \rightarrow \mathrm{P} ; \mathrm{B} \rightarrow \mathrm{Q} ; \mathrm{C} \rightarrow \mathrm{S} ; \mathrm{D} \rightarrow \mathrm{R}$ (c) $\mathrm{A} \rightarrow \mathrm{Q} ; \mathrm{B} \rightarrow \mathrm{P} ; \mathrm{C} \rightarrow \mathrm{R} ; \mathrm{D} \rightarrow \mathrm{S}$ (d) $\mathrm{A} \rightarrow \mathrm{Q} ; \mathrm{B} \rightarrow \mathrm{P} ; \mathrm{C} \rightarrow \mathrm{S} ; \mathrm{D} \rightarrow \mathrm{R}$

The deflection in a Galvanometer falls from 50 division to 20 when \(12 \Omega\) shunt is applied. The Galvanometer resistance is (a) \(18 \Omega\) (b) \(36 \Omega\) (c) \(24 \Omega\) (d) \(30 \Omega\)

The angles of dip at two places are \(30^{\circ}\) and \(45^{\circ}\). The ratio of horizontal components of earth's magnetic field at the two places will be (a) \(\sqrt{3}: \sqrt{2}\) (b) \(1: \sqrt{2}\) (c) \(1: 2\) (d) \(1: \sqrt{3}\)
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