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Chapter 13: Problem 1880

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

Short Answer

Expert verified
The number of turns in the coil is 50 (option c).

Step by step solution

01

Write down the given information and the formula

Given: Current, I = 10 A Radius, r = 10 cm = 0.1 m (converted to meters) Magnetic field at the center, B = \(\pi \times 10^{-3}\) T The formula for the magnetic field produced by a current-carrying circular coil at its center is: \[B = \frac{\mu_0 n I}{2r}\] where B is the magnetic field, \(\mu_0\) is the permeability of free space (\(4\pi \times 10^{-7} \mathrm{Tm/A}\)), n is the number of turns, I is the current, and r is the radius of the coil.
02

Solve for the number of turns (n)

We need to find the value of n. So, we will rearrange the formula for n and then substitute the given values. \[n = \frac{2rB}{\mu_0 I}\] Substituting the given values: \[n = \frac{2(0.1)(\pi \times 10^{-3})}{(4\pi \times 10^{-7})(10)}\]
03

Calculate the number of turns

Now, we will perform the calculation: \[n = \frac{(0.2\pi \times 10^{-3})}{(4\pi \times 10^{-7})(10)}\] \[n = \frac{0.2\pi \times 10^{-3}}{4\pi \times 10^{-6}}\] \[n = \frac{0.2}{4} \times 10^{3}\] \[n = 0.05 \times 10^{3}\] \[n = 50\] So, the number of turns in the coil is 50. The correct option is (c) 50.

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Most popular questions from this chapter

A Galvanometer coil has a resistance of \(15 \Omega\) and gives full scale deflection for a current of \(4 \mathrm{~mA}\). To convert it to an ammeter of range 0 to \(6 \mathrm{Amp}\) (a) \(10 \mathrm{~m} \Omega\) resistance is to be connected in parallel to the galvanometer. (b) \(10 \mathrm{~m} \Omega\) resistance is to be connected in series with the galvanometer. (c) \(0.1 \Omega\) resistance is to be connected in parallel to the galvanometer. (d) \(0.1 \Omega\) resistance is to be connected in series with the galvanometer.

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

A magnetic field \(B^{-}=\) Bo \(j \wedge\) exists in the region \(a

Two particles \(\mathrm{X}\) and \(\mathrm{Y}\) having equal charges, after being accelerated through the same potential difference, enter a region of uniform mag. field and describe circular path of radius \(\mathrm{r}_{1}\) and \(\mathrm{r}_{2}\) respectively. The ratio of mass of \(\mathrm{X}\) to that of \(\mathrm{Y}\) is (a) \(\sqrt{\left(r_{1} / \mathrm{r}_{2}\right)}\) (b) \(\left(\mathrm{r}_{2} / \mathrm{r}_{1}\right)\) (c) \(\left(\mathrm{r}_{1} / \mathrm{r}_{2}\right)^{2}\) (b) \(\left(\mathrm{r}_{1} / \mathrm{r}_{2}\right)\)

The strength of the magnetic field at a point \(\mathrm{y}\) near a long straight current carrying wire is \(\mathrm{B}\). The field at a distance \(\mathrm{y} / 2\) will be (a) B/2 (b) B \(/ 4\) (c) \(2 \mathrm{~B}\) (d) \(4 \mathrm{~B}\)
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