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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

A long horizontal wire " \(\mathrm{A}^{\prime \prime}\) carries a current of $50 \mathrm{Amp}$. It is rigidly fixed. Another small wire "B" is placed just above and parallel to " \(\mathrm{A}^{\prime \prime}\). The weight of wire- \(\mathrm{B}\) per unit length is \(75 \times 10^{-3}\) Newton/meter and carries a current of 25 Amp. Find the position of wire \(B\) from \(A\) so that wire \(B\) remains suspended due to magnetic repulsion. Also indicate the direction of current in \(B\) w.r.t. to \(A\).(a) \((1 / 2) \times 10^{-2} \mathrm{~m}\); in same direction (b) \((1 / 3) \times 10^{-2} \mathrm{~m}\); in mutually opposite direction (c) \((1 / 4) \times 10^{-2} \mathrm{~m}\); in same direction (d) \((1 / 5) \times 10^{-2} \mathrm{~m}\); in mutually opposite direction

A proton and an particle are projected with the same kinetic energy at right angles to the uniform mag. field. Which one of the following statements will be true. (a) The \(\alpha\) - particle will be bent in a circular path with a small radius that for the proton. (b) The radius of the path of the \(\alpha\) - particle will be greater than that of the proton. (c) The \(\alpha\) - particle and the proton will be bent in a circular path with the same radius. (d) The \(\alpha\) - particle and the proton will go through the field in a straight line.

The mag. field (B) at the centre of a circular coil of radius "a", through which a current I flows is (a) \(\mathrm{B} \propto \mathrm{c} \mathrm{a}\) (b) \(\mathrm{B} \propto(1 / \mathrm{I})\) (c) \(\mathrm{B} \propto \mathrm{I}\) (d) \(B \propto I^{2}\)

A long solenoid has 200 turns per \(\mathrm{cm}\) and carries a current of $2.5 \mathrm{Amp}$. The mag. field at its centre is tesla. (a) \(\pi \times 10^{-2}\) (b) \(2 \pi \times 10^{-2}\) (c) \(3 \pi \times 10^{-2}\) (d) \(4 \pi \times 10^{-2}\)

At a distance of \(10 \mathrm{~cm}\) from a long straight wire carrying current, the magnetic field is \(4 \times 10^{-2}\). At the distance of $40 \mathrm{~cm}$, the magnetic field will be Tesla. (a) \(1 \times 10^{-2}\) (b) \(2 \times \overline{10^{-2}}\) (c) \(8 \times 10^{-2}\) (d) \(16 \times 10^{-2}\)
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