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Chapter 13: Problem 1895

5: When the current flowing in a circular coil is doubled and the number of turns of the coil in it is halved, the magnetic field at its centre will become (a) Four times (b) Same (c) Half (d) Double

Short Answer

Expert verified
The magnetic field at the center of the circular coil will become double when the current is doubled and the number of turns is halved (\(B_2 = 2B_1\)).

Step by step solution

01

Formula for Magnetic Field at the Center of a Circular Coil

The formula for the magnetic field (B) at the center of a circular coil with radius R, carrying a current I, and having n turns is given by: \(B=\frac{\mu_0}{4\pi} \frac{2\pi nI}{R}\) Where \(\mu_0\) is the permeability of free space. 2. Calculate the initial magnetic field
02

Initial Magnetic Field

Let's denote the initial magnetic field as \(B_1\). Using the formula, we have: \(B_1=\frac{\mu_0}{4\pi} \frac{2\pi nI}{R}\) 3. Apply the given changes to the coil, current, and number of turns
03

Changes in Current and Number of Turns

The problem states that the current flowing in the coil is doubled, and the number of turns of the coil is halved. So, we can write the new current as \(2I\), and the new number of turns as \(\frac{n}{2}\). 4. Calculate the new magnetic field
04

New Magnetic Field

Let's denote the new magnetic field as \(B_2\). Using the formula with the new current and number of turns, we have: \(B_2=\frac{\mu_0}{4\pi} \frac{2\pi (\frac{n}{2})(2I)}{R}\) 5. Compare the initial and new magnetic field
05

Comparison of Magnetic Fields

To find the relationship between the initial and new magnetic field, let's divide \(B_2\) by \(B_1\): \(\frac{B_2}{B_1} = \frac{\frac{\mu_0}{4\pi} \frac{2\pi (\frac{n}{2})(2I)}{R}}{\frac{\mu_0}{4\pi} \frac{2\pi nI}{R}} = \frac{(\frac{n}{2})(2I)}{nI} = 2\) 6. Select the correct option
06

Correct Option

We found that the new magnetic field is two times the initial magnetic field (\(B_2 = 2B_1\)). Therefore, the correct answer is: (d) Double

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Most popular questions from this chapter

The time period of a freely suspended magnet is a 4 seconds. If it is broken in length into two equal parts and one part is suspended in the same way, then its time period will be (a) \(4 \mathrm{sec}\) (b) \(2 \mathrm{sec}\) (c) \(0.5 \mathrm{sec}\) (d) \(0.25 \mathrm{sec}\)

A coil in the shape of an equilateral triangle of side 115 suspended between the pole pieces of a permanent magnet such that \(\mathrm{B}^{-}\) is in plane of the coil. If due to a current \(\mathrm{I}\) in the triangle a torque \(\tau\) acts on it, the side 1 of the triangle is (a) \((2 / \sqrt{3})(\tau / \mathrm{BI})^{1 / 2}\) (b) \((2 / 3)(\tau / B I)\) (c) \(2[\tau /\\{\sqrt{(} 3) \mathrm{BI}\\}]^{1 / 2}\) (d) \((1 / \sqrt{3})(\tau / \mathrm{BI})\)

A 2 Mev proton is moving perpendicular to a uniform magnetic field of \(2.5\) tesla. The force on the proton is (a) \(3 \times 10^{-10} \mathrm{~N}\) (b) \(70.8 \times 10^{-11} \mathrm{~N}\) (c) \(3 \times 10^{-11} \mathrm{~N}\) (d) \(7.68 \times 10^{-12} \mathrm{~N}\)

An iron rod of length \(L\) and magnetic moment \(\mathrm{M}\) is bent in the form of a semicircle. Now its magnetic moment will be (a) \(\mathrm{M}\) (b) \((2 \mathrm{M} / \pi)\) (c) \((\mathrm{M} / \pi)\) (d) \(\mathrm{M} \pi\)

If a magnet of pole strength \(\mathrm{m}\) is divided into four parts such that the length and width of each part is half that of initial one, then the pole strength of each part will be (a) \((\mathrm{m} / 4)\) (b) \((\mathrm{m} / 2)\) (c) \((\mathrm{m} / 8)\) (d) \(4 \mathrm{~m}\)
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