An electron is accelerated by a potential difference of 12000, volts . It then enters a uniform magn

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4.Moving Charges and Magnetism

medium

An electron is accelerated by a potential difference of $12000\, volts$. It then enters a uniform magnetic field of ${10^{ - 3}}\,T$ applied perpendicular to the path of electron. Find the radius of path. Given mass of electron $ = 9 \times {10^{ - 31}}\,kg$ and charge on electron $ = 1.6 \times {10^{ - 19}}\,C$

$36.7\, m$

$36.7\, cm$

$3.67\, m$

$3.67\, cm$

Solution

(b) $r = \frac{{\sqrt {2mK} }}{{qB}} = \frac{1}{B}\sqrt {\frac{{2mV}}{q}} $
$ = \frac{1}{{{{10}^{ – 3}}}}\sqrt {\frac{{2 \times 9 \times {{10}^{ – 31}} \times 12000}}{{1.6 \times {{10}^{ – 19}}}}} $ = $0.367\, m = 36.7\, cm$

Standard 12

Physics

charged particle with charge $q$ enters a region of constant, uniform and mutually orthogonal fields $\vec E$ and $\vec B$ with a velocity $\vec v$ perpendicular to both $\vec E$ and $\vec B$ , and comes out without any change in magnitude or direction of $\vec v$ . Then

medium

(AIEEE-2007)

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hard

(IIT-2022)

The magnetic force acting on charged particle of charge $2\,\mu C$ in magnetic field of $2\, T$ acting in $y-$ direction , when the particle velocity is $\left( {2\hat i + 3\hat j} \right) \times {10^6}\,m{s^{ – 1}}$ is

medium

(AIEEE-2012)

A particle of specific charge $(q/m)$ is projected from the origin of coordinates with initial velocity $[ui – vj]$. Uniform electric magnetic fields exist in the region along the $+y$ direction, of magnitude $E$ and $B.$ The particle will definitely return to the origin once if

normal

A charge moving with velocity $v$ in $X$-direction is subjected to a field of magnetic induction in the negative $X$-direction. As a result, the charge will

easy

(AIPMT-1993)

Solution

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