Question 211:
easyUnder the influence of a uniform magnetic field, a charged particle moves with constant speed $V$ in a circle of radius $R$. The time period of rotation of the particle:
(2009)
The time period formula is $T = \frac{2\pi m}{qB}$. It depends only on mass, charge, and magnetic field, making it independent of both speed $V$ and radius $R$.
Question 212:
easyThe magnetic force acting on a charged particle of charge $-2\text{ }\mu\text{C}$ in a magnetic field of $2\text{ T}$ acting in $y$ direction, when the particle velocity is $(2\hat{i} + 3\hat{j}) \times 10^6\text{ ms}^{-1}$, is:
(2009)
Using $\vec{F} = q(\vec{v} \times \vec{B})$, substituting $q = -2 \times 10^{-6}\text{ C}$, $\vec{v} = (2\hat{i} + 3\hat{j}) \times 10^6\text{ ms}^{-1}$, and $\vec{B} = 2\hat{j}\text{ T}$ yields $-8\hat{k}\text{ N}$, i.e., $8\text{ N}$ in $-z$ direction.
Question 213:
easyAn electron moves in a circular orbit with a uniform speed $v$. It produces a magnetic field $B$ at the center of the circle. The radius of the circle is proportional to:
(2005)
Center field is $B = \frac{\mu_0 I}{2R}$ with $I = \frac{ev}{2\pi R}$, giving $B = \frac{\mu_0 ev}{4\pi R^2}$. Thus, radius $R$ is proportional to $\sqrt{v/B}$.
Question 214:
easyAn electron having mass ‘$m$’ and kinetic energy E enter in uniform magnetic field B perpendicularly, then its frequency will be:
(2001)
Cyclotron frequency is given by $f = \frac{qB}{2\pi m}$. For an electron, the charge is $e$, yielding $\frac{eB}{2\pi m}$.
Question 215:
easyA positively charged particle moving due East enters a region of uniform magnetic field directed vertically upwards. This particle will
(1997)
Magnetic force is always perpendicular to velocity, doing no work on the particle. Consequently, the speed remains constant while the particle moves in a circular path.
Question 216:
moderateA $10\text{ eV}$ electron is circulating in a plane at right angles to a uniform field at magnetic induction $10^{-4}\text{ Wb/m}^2$ ($= 1.0\text{ gauss}$), the orbital radius of electron is
(1996)
Using the radius formula $r = \frac{\sqrt{2mK}}{qB}$, substitute $K = 10text{ eV}$, $m = 9.1 times 10^{-31}text{ kg}$, and $B = 10^{-4}text{ Wb/m}^2$. Solving gives $r approx 11text{ cm}$.
Question 217:
easyA charge moving with velocity $v$ in X-direction is subjected to a field of magnetic induction in negative X-direction. As a result, the charge will
(1993)
When velocity and magnetic field are along the same or opposite directions, the angle between them is $180^circ$. The magnetic force is zero, so the charge remains unaffected.
Question 218:
easyA uniform magnetic field acts right angles to the direction of motion of electrons. As a result, the electron moves in a circular path of radius $2\text{ cm}$. If the speed of electrons is doubled, then, the radius of the circular path will be (1991)
The radius of the circular path is given by $$r = \frac{mv}{qB}$$. Since $r \propto v$, doubling the speed doubles the radius to $4.0\text{ cm}$.
Question 219:
easyA deuteron of kinetic energy $50\text{ keV}$ is describing a circular orbit of radius $0.5\text{ metre}$ in a plane perpendicular to magnetic field $B$. The kinetic energy of the proton that describes a circular, orbit of radius $0.5\text{ metre}$ in the same plane with the same $B$ is
(1991)
Using $r = \frac{\sqrt{2mK}}{qB}$, equating radii for deuteron and proton where $m_d = 2m_p$ yields $K_p = 2K_d = 100\text{ keV}$.
Question 220:
moderateWhen a proton is released from rest in a room, it starts with an initial acceleration $a_0$ towards west. When it is projected towards north with a speed $v_0$ it moves with an initial acceleration $3a_0$ toward west. The electric and magnetic fields in the room are:
(2013)
Electric field is $E = \frac{ma_0}{e}$ west from the initial acceleration. With velocity north, the magnetic force accounts for the extra $2a_0$ acceleration west, yielding a downward magnetic field of $B = \frac{2ma_0}{ev_0}$.