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Chapter 15: Problem 84

When a long wire carrying a steady current is bent into a circular coil of one turn, the magnetic induction at its centre is \(B\). When the same wire carrying the same current is bent to form a circular coil of \(n\) turns of a smaller radius, the magnetic induction at the centre will be (A) \(B / n\) (B) \(n B\) (C) \(B / n^{2}\) (D) \(n^{2} B\)

Short Answer

Expert verified
The magnetic induction at the center of the smaller circular coil with $n$ turns will be $n$ times the initial magnetic induction $B$. This is represented by option (B) $nB$.

Step by step solution

01

Understand the initial condition

Initially, the wire is bent into a circular coil of one turn. At the center of this coil, the magnetic induction is B. Since there is only one turn, we can represent this coil as having a magnetic induction of B1 = B.
02

Formulate the formula for magnetic induction at the center of a circular loop

According to Ampere's law, the magnetic induction at the center of a circular loop is given by: \[B = \frac{\mu_0 * I * n}{2 * R}\] where B = magnetic induction at the center of the loop 渭鈧� = permeability constant (4蟺 脳 10鈦烩伔 Tm/A) I = current through the loop n = number of turns R = radius of the loop
03

Calculate the initial magnetic induction B鈧�

Since the initial condition consists of one turn with a magnetic induction of B, we can rewrite the formula to represent the initial condition: \[B = \frac{\mu_0 * I * 1}{2 * R1}\] where R1 represents the radius of the circular coil in the initial condition.
04

Calculate the magnetic induction B鈧� for the coil with n turns and a smaller radius

Next, we will calculate the magnetic induction at the center of the circular coil of n turns with a smaller radius: \[B2 = \frac{\mu_0 * I * n}{2 * R2}\] where R2 represents the radius of the circular coil with n turns and R2 < R1.
05

Compare the two magnetic inductions B鈧� and B鈧�

We'll now compare the magnetic inductions B1 and B2: \[\frac{B1}{B2} = \frac{\frac{\mu_0 * I}{2 * R1}}{\frac{\mu_0 * I * n}{2 * R2}}\] Note that 渭鈧�, I, and 2 are constants and can be cancelled out: \[\frac{B1}{B2} = \frac{R2}{R1 * n}\]
06

Analyze the result

Since R2 < R1, the ratio of R2/R1 will be less than 1. Also, there is a factor of n on the denominator of the right-hand side. The closest option that fits this result is: (B) nB Hence, the magnetic induction at the center of the smaller circular coil with n turns will be n times the initial magnetic induction B.

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Most popular questions from this chapter

A particle of mass \(m\) and charge \(q\) is projected from origin with initial velocity \([u \hat{i}-v \hat{j}]\). Uniform electric and magnetic field exist in the region along \(+y\) direction of magnitudes \(E\) and \(B\), respectively. Then particle will definitively return to the origin if (A) \(\frac{v B}{2 \pi E}\) is an integer (B) \(\frac{\left(u^{2}+v^{2}\right)^{1 / 2}}{[B / \pi E]}\) is an integer (C) \(\frac{v B}{\pi E}\) is an integer (D) \(\frac{v B}{3 \pi E}\) is an integer

A magnetic needle is kept in a non-uniform magnetic field. It experiences \(\quad\) [2005] (A) neither a force nor a torque. (B) a torque but not a force. (C) a force but not a torque. (D) a force and a torque.

A long solenoid has 200 turns per \(\mathrm{cm}\) and carries a current \(i\). The magnetic field at its centre is \(6.28 \times 10^{-2}\) \(\mathrm{Wb} / \mathrm{m}^{2}\). Another long solenoid has 100 turns per \(\mathrm{cm}\) and it carries a current \(\frac{i}{3}\). The value of the magnetic field at its centre is (A) \(1.05 \times 10^{-4} \mathrm{~Wb} / \mathrm{m}^{2}\) (B) \(1.05 \times 10^{-2} \mathrm{~Wb} / \mathrm{m}^{2}\) (C) \(1.05 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^{2}\) (D) \(1.05 \times 10^{-3} \mathrm{~Wb} / \mathrm{m}^{2}\)

Two short bar magnets of length \(1 \mathrm{~cm}\) each have magnetic moments \(1.20 \mathrm{Am}^{2}\) and \(1.00 \mathrm{Am}^{2}\), respectively. They are placed on a horizontal table parallel to each other with their \(\mathrm{N}\) poles pointing towards the south. They have a common magnetic equator and are separated by a distance of \(20.0 \mathrm{~cm}\). The value of resultant horizontal magnetic induction at the midpoint \(O\) of the line joining their centres is close to [2013] (Horizontal component of earth's magnetic induction is \(\left.3.6 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^{2}\right)\) (A) \(2.56 \times 10^{-4} \mathrm{~Wb} / \mathrm{m}^{2}\) (B) \(3.50 \times 10^{-4} \mathrm{~Wb} / \mathrm{m}^{2}\) (C) \(5.80 \times 10^{-4} \mathrm{~Wb} / \mathrm{m}^{2}\) (D) \(3.6 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^{2}\)

A particle of charge \(-16 \times 10^{-18} \mathrm{C}\) moving with velocity \(10 \mathrm{~m} / \mathrm{s}\) along the \(x\)-axis enters a region where a magnetic field of induction \(B\) is along the \(y\)-axis and an electric field of magnitude \(10^{4} \mathrm{~V} / \mathrm{m}\) is along the negative \(z\)-axis. If the charged particle continues moving along the \(x\)-axis, the magnitude of \(B\) is (A) \(10^{3} \mathrm{~Wb} / \mathrm{m}^{2}\) (B) \(10^{5} \mathrm{~Wb} / \mathrm{m}^{2}\) (C) \(10^{16} \mathrm{~Wb} / \mathrm{m}^{2}\) (D) \(10^{-3} \mathrm{~Wb} / \mathrm{m}^{2}\)
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