Question 21:
difficultTwo circular coils X and Y have equal number of turns and carry equal currents in the same sense and subtend same solid angle at point O. If the smaller coil X is midway between O and Y, then if we represent the magnetic induction due to bigger coil Y at O as BY and that due to smaller coil X at O as Bx :
Question 22:
difficultA thin flexible wire of length L is connected to two adjacent fixed points carries a current I in the clockwise direction, as shown in the figure. When system is put in a uniform magnetic field of strength B going into the plane of paper, the wire takes the shape of a circle. The tension in the wire is :
Question 23:
moderateA helium nucleus is moving in a circular path of radius \(0.8\text{ m}\). If it takes \(2\text{ sec}\) to complete one revolution, the magnetic field produced at the centre of the circle is:
Current is \(I = \frac{q}{T} = \frac{2e}{2} = e = 1.6 \times 10^{-19}\text{ A}\). The magnetic field at the centre is \(B = \frac{\mu_0 I}{2R} = \frac{\mu_0 (1.6 \times 10^{-19})}{2(0.8)} = \mu_0 \times 10^{-19}\text{ T}\).
Question 24:
easyConsider two long solenoids \( A \) and \( B \) having length \( 2L \) and \( 3L \) and number of loop as \( N \) and \( 2N \) respectively. If both have same current then ratio of magnetic field inside \( A \) to that of the \( B \) will be
The magnetic field inside a solenoid is given by \( B = \mu_0 \frac{N}{L} I \). Calculating the ratio: \( \frac{B_A}{B_B} = \frac{N_A / L_A}{N_B / L_B} = \frac{N / 2L}{2N / 3L} = \frac{3}{4} \).
Question 25:
moderateA long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the center of the loop is B. It is then bent into a circular coil of n turns. The magnetic field at the center of this coil of n turns will be:
(2016 – II)
For a single turn loop, radius $R = \frac{L}{2\pi}$ and $B = \frac{\mu_0 i}{2R}$. When bent into $n$ turns, the new radius is $R' = \frac{R}{n}$. The magnetic field becomes $B' = \frac{\mu_0 n i}{2R'} = n^2 B$.
Question 26:
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 27:
moderateTwo similar coils of radius R are lying concentrically with their planes at right angles to each other. The currents flowing in them are I and 2I, respectively. The resultant-magnetic field induction at the center will be:
(2012 Pre)
The magnetic fields due to the two perpendicular coils are $B_1 = \frac{\mu_0 I}{2R}$ and $B_2 = \frac{\mu_0 (2I)}{2R} = \frac{\mu_0 I}{R}$. Since their planes are at right angles, the fields are perpendicular, so $B_{\text{res}} = \sqrt{B_1^2 + B_2^2} = \frac{\sqrt{5}\mu_0 I}{2R}$.
Question 28:
moderateCharge q is uniformly spread on a thin ring of radius R. The ring rotates about its axis with a uniform frequency f Hz. The magnitude of magnetic induction at the center of the ring is:
(2011 Mains)
The equivalent current is $i = qf$. The magnetic induction at the center of a circular ring of radius $R$ carrying current $i$ is $B = \frac{\mu_0 i}{2R} = \frac{\mu_0 q f}{2R}$.
Question 29:
moderateThe magnetic field of given length of wire for single turn coil at its centre is ‘B’ then its value for two turns coil for the same wire is:
(2002)
For a wire of length $L$, radius for single turn is $R = L/(2\pi)$ and for two turns is $R' = L/(4\pi) = R/2$. Magnetic field $B = \frac{\mu_0 N I}{2 R}$. For 2 turns, $B' = \frac{\mu_0 (2) I}{2 (R/2)} = 4 \left(\frac{\mu_0 I}{2R}\right) = 4B$.