Question 111:
easyA long solenoid carrying a current produces a magnetic field $B$ along its axis. If the current is doubled and the number of turns per cm is halved, the new value of the magnetic field is:
(2003)
The magnetic field of a solenoid is $B = mu_0 n I$. When the current is doubled ($I' = 2I$) and turns per unit length are halved ($n' = n/2$), the new magnetic field is $B' = mu_0 (n/2)(2I) = mu_0 n I = B$.
Question 112:
easyAn electron is moving in a circular path under the influence of a transverse magnetic field of $3.57 \times 10^{-2}\text{ T}$. If the value of $e/m$ is $1.76 \times 10^{11}\text{ C/kg}$, the frequency of revolution of the electron is:
(2016 – II)
The frequency of revolution is $f = \frac{eB}{2\pi m}$. Substituting values yields $f = \frac{1.76 \times 10^{11} \times 3.57 \times 10^{-2}}{2\pi} \approx 6.28\text{ MHz}$.
Question 113:
easyA proton and an alpha particle both enter a region of uniform magnetic field, $B$, moving at right angles to the field $B$. If the radius of circular orbits for both the particles is equal and the kinetic energy acquired by proton is $1\text{ MeV}$, the energy acquired by the alpha particle will be:
(2015 Pre)
Radius is $r = \frac{\sqrt{2mK}}{qB}$. For equal radii, $K \propto \frac{q^2}{m}$. Since $q_\alpha = 2q_p$ and $m_\alpha = 4m_p$, the kinetic energy remains $1\text{ MeV}$.
Question 114:
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 115:
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 116:
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 117:
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 118:
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 119:
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 120:
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 $K_p = 2K_d = 100\text{ keV}$.