Chapter 13: Problem 1959
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
Chapter 13: Problem 1959
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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Get started for freeA conducting circular loop of radius a carries a constant current I. It is placed in a uniform magnetic field \(\mathrm{B}^{-}\), such that \(\mathrm{B}^{-}\) is perpendicular to the plane of the Loop. The magnetic force acting on the Loop is (a) \(\mathrm{B}^{-} \operatorname{Ir}\) (b) \(\mathrm{B}^{-} \mathrm{I} \pi \mathrm{r}^{2}\) (c) Zero (d) BI \((2 \pi \mathrm{r})\)
Force between two identical bar magnets whose centers are I meter apart is \(4.8 \mathrm{~N}\), when their axes are in the same line. If separation is increased to \(2 r\), the force between them is reduced to (a) \(2.4 \mathrm{~N}\) (b) \(1.2 \mathrm{~N}\) (c) \(0.6 \mathrm{~N}\) (d) \(0.3 \mathrm{~N}\)
A thin magnetic needle oscillates in a horizontal plane with a period \(\mathrm{T}\). It is broken into n equal parts. The time period of each part will be (a) \(\mathrm{T}\) (b) \(\mathrm{n}^{2} \mathrm{~T}\) (c) \((\mathrm{T} / \mathrm{n})\) (d) \(\left(\mathrm{T} / \mathrm{n}^{2}\right)\)
In a H-atom, an electron moves in a circular orbit of radius $5.2 \times 10^{-11}\( meter and produces a mag. field of \)12.56$ Tesla at its nucleus. The current produced by the motion of the electron will be (a) \(6.53 \times 10^{-3}\) (b) \(13.25 \times 10^{-10}\) (c) \(9.6 \times 10^{6}\) (d) \(1.04 \times 10^{-3}\)
A straight wire of mass \(200 \mathrm{gm}\) and length \(1.5\) meter carries a current of 2 Amp. It is suspended in mid-air by a uniform horizontal magnetic field B. [take \(\left.\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right]\). The \(\mathrm{B}\) is (a) \(\overline{(2 / 3) \text { tes } 1 a}\) (b) \((3 / 2)\) tesla (c) \((20 / 3)\) tesla (d) (3/20) tesla
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