C”eqpfwevqt”nkgu”cnqpi”vjg”|/czku”cv”- £< 307 | 307 o"cpf ecttkgu"c"hkzgf"ewttgpv"qh"3202"C"kp" | -cŠ "fktgevkqp"*ugg"hkiwtg+0 Hqt"c"hkgnf" 6 204z { D 502 32 g cŠ - - = ´ r "V."hkpf"vjg"rqygt"tgswktgf vq"oqxg"vjg"eqpfwevqt"cv"eqpuvcpv"urggf"vq"z"?"402"o. {"?"2"o"kp" 5 7 32 u0 - ´ "Cuuwog"rctcnngn"oqvkqp"cnqpi"vjg z/czku0


A conductor lies along the $z$-axis at $-1.5 \leq z<1.5 \mathrm{~m}$ and carries a fixed current of $10.0$ A in $-\hat{a}_{z}$ direction (see figure). For a field $\overrightarrow{\mathrm{B}}=3.0 \times 10^{-4} \mathrm{e}^{-0.2 \mathrm{x}} \hat{\mathrm{a}}_{\mathrm{y}} \mathrm{T}$, find the power required to move the conductor at constant speed to $\mathrm{x}=2.0 \mathrm{~m}$, $\mathrm{y}=0 \mathrm{~m}$ in $5 \times 10^{-3} \mathrm{~s}$. Assume parallel motion along the $x$-axis.

  1. $1.57 \mathrm{~W}$

  2. $2.97 \mathrm{~W}$

  3. $14.85 \mathrm{~W}$

  4. $29.7 \mathrm{~W}$

Correct Option: 3

JEE Main Previous Year 1 Question of JEE Main from Physics Moving Charges and Magnetism chapter.
JEE Main Previous Year 2014


Related Questions

  • asdf

    View Solution

  • questions

    View Solution

  • An electron is moving along $+x$ direction with a velocity of $6 \times 10^{6} \mathrm{~ms}^{-1}$. It enters a region of uniform electric field of $300 \mathrm{~V} / \mathrm{cm}$ pointing along $+y$ direction. The magnitude and direction of the magnetic field set up in this region such that the electron keeps moving along the $x$ direction will be:

    View Solution

  • A particle of charge $q$ and mass $m$ is moving with a velocity $-v \hat{i}(v \neq 0)$ towards a large screen placed in the $\mathrm{Y}-\mathrm{Z}$ plane at a distance $d$. If there is a magnetic field $\vec{B}=B_{0} \hat{k}$, the minimum value of $v$ for which the particle will not hit the screen is:

    View Solution

  • A charged particle carrying charge $1 \mu \mathrm{C}$ is moving with velocity $(2 \hat{i}+3 \hat{j}+4 \hat{k}) \mathrm{ms}^{-1}$. If an external magnetic field of $(5 \hat{i}+3 \hat{j}-6 \hat{k}) \times 10^{-3} \mathrm{~T}$ exists in the region where the particle is moving then the force on the particle is $\vec{F} \times 10^{-9} \mathrm{~N}$. The vector $\vec{F}$ is :

    View Solution

  • A beam of protons with speed $4 \times 10^{5} \mathrm{~ms}^{-1}$ enters a uniform magnetic field of $0.3 \mathrm{~T}$ at an angle of $60^{\circ}$ to the magnetic field. The pitch of the resulting helical path of protons is close to : (Mass of the proton $=1.67 \times 10^{-27} \mathrm{~kg}$, charge of the proton $=1.69 \times 10^{-19} \mathrm{C}$ )

    View Solution

  • The figure shows a region of length ‘ $l$ ‘ with a uniform magnetic field of $0.3 \mathrm{~T}$ in it and a proton entering the region with velocity $4 \times 10^{5} \mathrm{~ms}^{-1}$ making an angle $60^{\circ}$ with the field. If the proton completes 10 revolution by the time it cross the region shown, ‘ $l$ ‘ is close to (mass of proton $=1.67 \times 10^{-27} \mathrm{~kg}$, charge of the proton $=1.6 \times 10^{-19} \mathrm{C}$ )

    View Solution

  • Proton with kinetic energy of $1 \mathrm{MeV}$ moves from south to north. It gets an acceleration of $10^{12} \mathrm{~m} / \mathrm{s}^{2}$ by an applied magnetic field (west to east). The value of magnetic field: (Rest mass of proton is $1.6 \times 10^{-27} \mathrm{~kg}$ )

    View Solution

  • A particle having the same charge as of electron moves in a circular path of radius $0.5 \mathrm{~cm}$ under the influence of a magnetic field of $0.5 \mathrm{~T}$. If an electric field of $100 \mathrm{~V} / \mathrm{m}$ makes it to move in a straight path then the mass of the particle is (Given charge of electron $=1.6 \times 10^{-19} \mathrm{C}$ )

    View Solution

  • An electron, moving along the $x$-axis with an initial energy of $100 \mathrm{eV}$, enters a region of magnetic field $\vec{B}=\left(1.5 \times 10^{-3} \mathrm{~T}\right) \hat{k}$ at $\mathrm{S}$ (see figure). The field extends between $x=0$ and $x=$ $2 \mathrm{~cm}$. The electron is detected at the point Q on a screen placed $8 \mathrm{~cm}$ away from the point $\mathrm{S}$. The distance $d$ between $\mathrm{P}$ and $\mathrm{Q}$ (on the screen) is :

    (Electron’s charge $=1.6 \times 10^{-19} \mathrm{C}$, mass of electron $=9.1 \times 10^{-31} \mathrm{~kg}$ )

    View Solution

Leave a Reply

Your email address will not be published.

error: Content is protected !!
Download App