Question 201:
moderateAn electron moving in a circular orbit of radius r makes n rotations per second. The magnetic field produced at the center has magnitude:
(2015)
The current produced by the revolving electron is $i = qf = ne$. The magnetic field at the center of a circular loop of radius $r$ carrying current $i$ is $B = \frac{\mu_0 i}{2r} = \frac{\mu_0 n e}{2r}$.
Question 202:
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 203:
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 204:
easyA long solenoid of radius $1\text{ mm}$ has $100$ turns per mm. If $1\text{ A}$ current flows in the solenoid, the magnetic field strength at the centre of the solenoid is:
(2022)
The magnetic field inside a long solenoid is given by $B = \mu_0 n I$. Given $n = 100\text{ turns/mm} = 10^5\text{ turns/m}$ and $I = 1\text{ A}$, substituting the values yields $B = (4\pi \times 10^{-7}) \times 10^5 \times 1 = 12.56 \times 10^{-2}\text{ T}$.
Question 205:
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 206:
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 207:
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 208:
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 209:
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 210:
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 $1\text{ MeV}$.