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Chapter 13: Problem 1892

Two similar coils are kept mutually perpendicular such that their centers co- inside. At the centre, find the ratio of the mag. field due to one coil and the resultant magnetic field by both coils, if the same current is flown. (a) \(1: \sqrt{2}\) (b) \(1: 2\) (c) \(2: 1\) (d) \(\sqrt{3}: 1\)

Short Answer

Expert verified
The ratio of the magnetic field due to one coil and the resultant magnetic field by both coils is 1:√2. The correct answer is (a) 1:\(\sqrt{2}\).

Step by step solution

01

Determine the Magnetic Field due to One Coil

First, we need to find the magnetic field due to one coil at the center. Since the coils are similar and the current is the same in both coils, the magnetic field due to one coil will be equal to the magnetic field of the other coil. Let's call this magnetic field B1. For a circular coil with current I and N turns, the magnetic field B at the center can be calculated using the formula: \[B = \frac{\mu_0}{4 \pi}\frac{2 \pi NI}{R}\] Since we're dealing with only one coil here, N = 1. Now, the equation becomes: \[B1 = \frac{\mu_0 I}{2R}\]
02

Determine the Magnetic Field due to Both Coils

Now, we will find the magnetic field due to both coils at the center. Since the coils are mutually perpendicular, their magnetic fields will be orthogonal to each other, and the resultant magnetic field can be calculated using the Pythagorean theorem. Thus, the magnetic field due to both coils, B_total, can be found as: \[B_{total} = \sqrt{B1^2 + B2^2}\] Since the coils are similar and have the same current, B1 = B2, so: \[B_{total} = \sqrt{B1^2 + B1^2} = B1\sqrt{2}\]
03

Calculate the Ratio of Magnetic Fields

We now have the magnetic field due to one coil (B1) and the total magnetic field due to both coils (B_total). To find the required ratio, we divide the total magnetic field by the magnetic field due to one coil: \[\frac{B1}{B_{total}} = \frac{B1}{B1\sqrt{2}} = \frac{1}{\sqrt{2}}\] So, the ratio of the magnetic field due to one coil and the resultant magnetic field by both coils is 1:√2. The correct answer is (a) 1:\(\sqrt{2}\).

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Most popular questions from this chapter

The mag. 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 the LOOP. (a) \(250 \mu \mathrm{T}\) (b) \(150 \mu \mathrm{T}\) (c) \(125 \mu \mathrm{T}\) (d) \(75 \mu \mathrm{T}\)

A magnetic field existing in a region is given by $\mathrm{B}^{-}=\mathrm{B}_{0}[1+(\mathrm{x} / \ell)] \mathrm{k} \wedge . \mathrm{A}\( square loop of side \)\ell$ and carrying current I is placed with edges (sides) parallel to \(\mathrm{X}-\mathrm{Y}\) axis. The magnitude of the net magnetic force experienced by the Loop is (a) \(2 \mathrm{~B}_{0} \overline{\mathrm{I} \ell}\) (b) \((1 / 2) B_{0} I \ell\) (c) \(\mathrm{B}_{\circ} \mathrm{I} \ell\) (d) BI\ell

Two iclentical short bar magnets, each having magnetic moment \(\mathrm{M}\) are placed a distance of \(2 \mathrm{~d}\) apart with axes perpendicular to each other in a horizontal plane. The magnetic induction at a point midway between them is. (a) $\sqrt{2}\left(\mu_{0} / 4 \pi\right)\left(\mathrm{M} / \mathrm{d}^{3}\right)$ (b) $\sqrt{3}\left(\mu_{0} / 4 \pi\right)\left(\mathrm{M} / \mathrm{d}^{3}\right)$ (c) $\sqrt{4}\left(\mu_{0} / 4 \pi\right)\left(\mathrm{M} / \mathrm{d}^{3}\right)$ (d) $\sqrt{5}\left(\mu_{0} / 4 \pi\right)\left(\mathrm{M} / \mathrm{d}^{3}\right)$

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) \(1.04 \times 10^{-3}\)

0: Due to 10 Amp of current flowing in a circular coil of \(10 \mathrm{~cm}\) radius, the mag. field produced at its centre is \(\pi \times 10^{-3}\) Tesla. The number of turns in the coil will be (a) 5000 (b) 100 (c) 50 (d) 25
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