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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

A small bar magnet has a magnetic moment $1.2 \mathrm{~A} \cdot \mathrm{m}^{2}\(. The magnetic field at a distance \)0.1 \mathrm{~m}$ on its axis will be tesla. (a) \(1.2 \times 10^{-4}\) (b) \(2.4 \times 10^{-4}\) (c) \(2.4 \times 10^{4}\) (d) \(1.2 \times 10^{4}\)

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) \(4 \pi \times 10^{-2}\)

In a mass spectrometer used for measuring the masses of ions, the ions are initially accelerated by an ele. potential \(\mathrm{V}\) and then made to describe semicircular paths of radius \(\mathrm{r}\) using a magnetic field \(\mathrm{B}\). If \(\mathrm{V}\) and \(\mathrm{B}\) are kept constant, the ratio [(Charge on the ion) / (mass of the ion)] will be proportional to. (a) \(\left(1 / r^{2}\right)\) (b) \(r^{2}\) (c) \(\mathrm{r}\) (d) \((1 / \mathrm{r})\)

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

A He nucleus makes a full rotation in a circle of radius \(0.8\) meter in $2 \mathrm{sec}\(. The value of the mag. field \)\mathrm{B}$ at the centre of the circle will be \(\quad\) Tesla. (a) \(\left(10^{-19} / \mu_{0}\right)\) (b) \(10^{-19} \mu_{0}\) (c) \(2 \times 10^{-10} \mathrm{H}_{0}\) (d) \(\left[\left(2 \times 10^{-10}\right) / \mu_{0}\right]\)
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