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Chapter 13: Problem 1893

A long wire carr1es a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is \(\mathrm{B}\). It is then bent into a circular Loop of n turns. The magnetic field at the centre of the coil for same current will be. (a) \(\mathrm{nB}\) (b) \(\mathrm{n}^{2} \mathrm{~B}\) (c) \(2 \mathrm{nB}\) (d) \(2 \mathrm{n}^{2} \mathrm{~B}\)

Short Answer

Expert verified
The magnetic field at the center of the coil for the same current with 'n' turns will be (a) nB.

Step by step solution

01

Recall relationship between magnetic field and coil attributes

The magnetic field at the center of a circular loop is given by the formula: \[B = \frac{\mu_0 * I * R}{2 R^2}\] Where: - \(B\) is the magnetic field at the center of the loop, - \(\mu_0\) is the permeability of free space, - \(I\) is the current through the wire, - \(R\) is the radius of the circular loop. In this case, when the coil has one turn, the magnetic field is given as \(B\).
02

Find the expression for magnetic field with n turns

Now we need to extend this expression for a coil with 'n' turns. The magnetic field produced by each turn is the same, given by the formula in step 1. Since the loops are wound on top of each other, the magnetic field at the center will add up. Thus, the total magnetic field at the center of the coil with 'n' turns would be: \[B_{total} = n * B\] Where: - \(B_{total}\) is the total magnetic field at the center, - \(n\) is the number of turns. Now let's find the correct answer from the given options.
03

Match the result with given options

Comparing our derived expression with the given options, we can see that: \[B_{total} = n * B\] Corresponds to option (a) nB. So, the magnetic field at the center of the coil for the same current with 'n' turns will be: (a) nB

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Most popular questions from this chapter

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 unit of ele. current "AMPERE" is the current which when flowing through each of two parallel wires spaced 1 meter apart in vacuum and of infinite length will give rise to a force between them equal to $\mathrm{N} / \mathrm{m}$ (a) 1 (b) \(2 \times 10^{-7}\) (c) \(1 \times 10^{-2}\) (d) \(4 \pi \times 10^{-7}\)

An electron having mass \(9 \times 10^{-31} \mathrm{~kg}\), charge $1.6 \times 10^{-19} \mathrm{C}\( and moving with a velocity of \)10^{6} \mathrm{~m} / \mathrm{s}$ enters a region where mag. field exists. If it describes a circle of radius \(0.10 \mathrm{~m}\), the intensity of magnetic field must be Tesla (a) \(1.8 \times 10^{-4}\) (b) \(5.6 \times \overline{10^{-5}}\) (c) \(14.4 \times 10^{-5}\) (d) \(1.3 \times 10^{-6}\)

A solenoid of \(1.5\) meter length and \(4 \mathrm{~cm}\) diameter possesses 10 turn per \(\mathrm{cm}\). A current of \(5 \mathrm{Amp}\) is flowing through it. The magnetic induction at axis inside the solenoid is (a) \(2 \pi \times 10^{-3} \mathrm{~T}\) (b) \(2 \pi \times 10^{-5} \mathrm{~T}\) (c) \(2 \pi \times 10^{-2} \mathrm{G}\) (d) \(2 \pi \times 10^{-5} \mathrm{G}\)

The magnetism of magnet is due to (a) The spin motion of electron (b) Earth (c) Pressure inside the earth core region (d) Cosmic rays
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