Question 71:
moderateA long solenoid of $50text{ cm}$ length having $100$ turns carries a current of $2.5text{ A}$. The magnetic field at the centre of the solenoid is:Â ($\mu_0 = 4\pi \times 10^{-7}\text{ T m A}^{-1}$)
(2020)
The number of turns per unit length is $n = \frac{N}{L} = \frac{100}{0.5\text{ m}} = 200\text{ turns/m}$. Using the formula $B = \mu_0 n I$, we substitute $\mu_0 = 4\pi \times 10^{-7}$, $n = 200$, and $I = 2.5\text{ A}$ to get $B = 6.28 \times 10^{-4}\text{ T}$.
Question 72:
moderateFrom Ampere’s circuital law for a long straight wire of circular cross section carrying a steady current, the variation of magnetic field in the inside and outside region of the wire is:
(2022)
Inside a uniform cylindrical wire, the magnetic field is directly proportional to the radius ($B \propto r$), increasing linearly. Outside the wire, the magnetic field is inversely proportional to the distance ($B \propto 1/r$). Therefore, option D is the correct choice.
Question 73:
moderateA long straight wire of radius $a$ carries a steady current $I$. The current is uniformly distributed over its cross-section. The ratio of the magnetic fields $B$ and $B’$ at radial distances $\frac{a}{2}$ and $2a$ respectively, from the axis of the wire is:
(2016-1)
The magnetic field inside the wire at distance $r = a/2$ is given by $B = \frac{\mu_0 I r}{2\pi a^2}$. The magnetic field outside at $r' = 2a$ is $B' = \frac{\mu_0 I}{2\pi r'}$. Evaluating both gives equal magnitudes, so the ratio $B / B'$ is equal to $1$.
Question 74:
moderateA charge having $q/m$ equal to $10^8\text{ C/kg}$ and with velocity $3 \times 10^5\text{ m/s}$ enters into a uniform magnetic field $B = 0.3\text{ tesla}$ at an angle $30^\circ$ with direction of field. Then radius of curvature will be:
(2000)
Radius is $r = \frac{vsin\theta}{(q/m)B}$. Substituting values yields $r = \frac{3 \times 10^5 sin(30^\circ)}{10^8 \times 0.3} = 0.005\text{ m} = 0.5\text{ cm}$.
Question 75:
moderateA charge having $q/m$ equal to $10^8\text{ C/kg}$ and with velocity $3 \times 10^5\text{ m/s}$ enters into a uniform magnetic field $B = 0.3\text{ tesla}$ at an angle $30^\circ$ with direction of field. Then radius of curvature will be:
(2000)
Radius is $r = \frac{v\sin\theta}{(q/m)B}$. Substituting values yields $r = \frac{3 \times 10^5 \sin(30^\circ)}{10^8 \times 0.3} = 0.005\text{ m} = 0.5\text{ cm}$.
Question 76:
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 77:
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 $B = \frac{2ma_0}{ev_0}$.
Question 78:
moderateAn alternating electric field, of frequency $\nu$, is applied across the dees (radius $= R$) of a cyclotron that is being used to accelerate protons ($\text{mass} = m$). The operating magnetic field ($B$) used in the cyclotron and the kinetic energy ($K$) of the proton beam, produced by it, are given by:
(2012 Pre)
Cyclotron frequency is $\nu = \frac{eB}{2\pi m}$, giving $B = \frac{2\pi m\nu}{e}$. Maximum kinetic energy is $K = \frac{e^2B^2R^2}{2m} = 2m\pi^2\nu^2R^2$.
Question 79:
moderateA beam of cathode rays is subjected to crossed Electric (E) and Magnetic field (B). The fields are adjusted such that the beam is not deflected. The specific charge of the cathode rays is given by (where V is the potential difference between cathode and anode):
(2010 Pre)
For undeflected motion, velocity is $v = \frac{E}{B}$.
Equating kinetic energy to electrical work gives $\frac{1}{2}mv^2 = eV$.
Substituting velocity yields specific charge $\frac{e}{m} = \frac{E^2}{2VB^2}$.
Question 80:
moderateA charge ‘q’ moves in a region where electric field and magnetic field both exist, then force on it is:
(2002)
The total electromagnetic force on a moving charge in both electric and magnetic fields is the Lorentz force.
It is the vector sum of the electric force $q\vec{E}$ and the magnetic force $q(\vec{V} \times \vec{B})$.