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Chapter 15: Problem 70

A point charge of \(0.1 \mathrm{C}\) is placed on the circumference of a non- conducting ring of radius \(1 \mathrm{~m}\) which is rotating with a constant angular acceleration of \(1 \mathrm{rad} / \mathrm{s}^{2}\). If ring starts its motion at \(t=0\), the magnetic field at the centre of the ring, at \(t=10 \mathrm{~s}\), is (A) \(10^{-6} \mathrm{~T}\) (B) \(10^{-7} \mathrm{~T}\) (C) \(10^{-8} \mathrm{~T}\) (D) \(10^{7} \mathrm{~T}\)

Short Answer

Expert verified
The magnetic field at the center of the ring at t = 10s is approximately \(10^{-7} \mathrm{~T}\).

Step by step solution

01

Calculate the angular velocity

We are given the angular acceleration (伪) and time (t). We can calculate the angular velocity (蠅) at time t using the formula: 蠅 = 伪 * t Given 伪 = 1 rad/s虏 and t = 10s, we have: 蠅 = 1 * 10 蠅 = 10 rad/s
02

Determine the linear velocity

Now, we will calculate the linear velocity (v) of the point charge on the circumference of the ring, given by the formula: v = 蠅 * r Given 蠅 = 10 rad/s and r = 1 m, we have: v = 10 * 1 v = 10 m/s
03

Apply the Biot-Savart law

The Biot-Savart law can be applied to find the magnetic field (B) generated by the moving charge: B = \(\frac{\mu_0 qv}{2 \pi r}\) Where \(\mu_0\) is the permeability of free space, q is the charge, v is the linear velocity, and r is the radius. Given \(\mu_0 = 4 \pi 脳 10^{-7} \mathrm{T.m/A}\), q = 0.1C, v = 10 m/s, and r = 1m, we have: B = \(\frac{(4 \pi 脳 10^{-7})(0.1)(10)}{2 \pi (1)}\)
04

Calculate the magnetic field

Now, calculate the magnetic field (B): B = \(2 \times 10^{-7} \mathrm{T}\) From the given options, this value is closest to option (B) \(10^{-7} \mathrm{~T}\). So, the correct answer is (B).

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Most popular questions from this chapter

Assertion: A conducting circular disc of radius \(R\) rotates about its own axis with angular velocity \(\omega\) in uniform magnetic field \(B_{0}\) along axis of the disc then no EMF is induced in the disc. Reason: Whenever a conductor cuts across magnetic lines of flux, an EMF is induced in the conductor. (A) \(\mathrm{A}\) (B) \(\mathrm{B}\) (C) \(\overline{\mathrm{C}}\) (D) D

A conductor \(A B\) carries a current \(i\) in a magnetic field \(\vec{B}\). If \(\overrightarrow{A B}=\vec{r}\) and the force on the conductor is \(\vec{F}\) (A) \(\vec{F}\) does not depend on the shape of \(A B\). (B) \(\vec{F}=i(\vec{r} \times \vec{B})\). (C) \(\vec{F}=i(\vec{B} \times \vec{r})\). (D) \(|\vec{F}|=i(\vec{r} \cdot \vec{B})\).

Two long conductors, separated by a distance \(d\), carry current \(I_{1}\) and \(I_{2}\) in the same direction. They exert a force \(F\) on each other. Now the current in one of them is doubles and its direction is reversed. The distance is also increased to \(3 d\). The new value of the force between them is (A) \(-\frac{2 F}{3}\) (B) \(\frac{F}{3}\) (C) \(-2 F\) (D) \(-\frac{F}{3}\)

Two circular coils of radii \(5 \mathrm{~cm}\) and \(10 \mathrm{~cm}\) carry equal currents of \(2 \mathrm{~A}\). The coils have 50 and 100 turns, respectively, and are placed in such a way that their planes and their centres coincide. Magnitude of magnetic field at the common centre of coils is, (A) \(8 \pi \times 10^{-4} \mathrm{~T}\) if currents in the coil are in same direction. (B) \(4 \pi \times 10^{-4} \mathrm{~T}\) if currents in the coil are in opposite direction. (C) zero, if currents in the coils are in opposite direction. (D) \(8 \pi \times 10^{-4} \mathrm{~T}\) if currents in the coil are in opposite direction.

A particle of mass \(m\) and charge \(q\) moves with a constant velocity \(v\) along the positive \(x\) direction. It enters a region containing a uniform magnetic field \(B\) directed along the negative \(z\)-direction, extending from \(x=a\) to \(x=b .\) The minimum value of \(v\) required so that the particle can just enter the region \(x>b\) is (A) \(q b B / m\) (B) \(q(b-a) B / m\) (C) \(q a B / m\) (D) \(q(b+a) B / 2 m\)
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