Question 1

Question 2

A rectangular loop of length  l  and breadth  b  is placed at a distance of  x  from an infinitely long wire carrying current  i  such that the direction of current is parallel to breadth. If the loop moves away from the current wire in a direction perpendicular to it with a velocity  v , the magnitude of the e.m.f. in the loop is ( \mu _{0}  permeability of free space)

Accepted Answer

Consider a strip of width  dy  at  y  distance from the wirefor infinite long wire B=\dfrac{\mu_o}{4\pi }\dfrac{2I}{y} area of strips =bdy d\phi =2I\dfrac{\mu_o}{4\pi}\dfrac{b}{y}dy \phi =\int_{x}^{x+l}2I\dfrac{\mu_o}{4\pi}\dfrac{b}{y}dy \phi=\dfrac{2I\mu_ob}{4\pi}ln\dfrac{x+l}{x} now Induced emf   =\dfrac{d\phi}{dt}=\dfrac{I\mu_ob}{2\pi}(\dfrac{x}{x+l})(-\dfrac{l}{x^2})\dfrac{dx}{dt} Magnitude of emf   =\dfrac{I\mu_obl}{2\pi}(\dfrac{v}{x(x+l)})

Question 3

Question 4

Question 5

Question 6

When the acceleration of rod is zero, the charge on capacitor is :

Accepted Answer

At any instant  t,  the charge on capacitor q, velocity of rod  v  and the current  I  through rod are as shown 	herefore m\frac{dv}{dt}=BIL =B(-\dfrac{dq}{dt})L....(1) \Rightarrow \int_{0}^{v}m dv=-\int_{Q_{0}}^{q}BL dq solving we get  q,\  q=Q_0-\dfrac{Mv}{BL} ................(2) Also \dfrac{q}{C}=BLV+IR=BLv-R\dfrac{dq}{BL}..........(3) From equations  (1)  and  (3) \dfrac{q}{C}=Blv+\dfrac{mR}{BL}\dfrac{dv}{dt}.................(4) from (4) when  \dfrac{dv}{dt}=0\Rightarrow \dfrac{q}{C}=Blv....(5) From (2) and (5). At instant acceleration is zero v=\dfrac{Q_0LB}{M+B^2L^2C}  and  q=\dfrac{B^2L^2CQ_0}{M+B^2L^2C}

Question 7

A thin insulated wire forms a plane spiral of

N = 100

tight turns carrying a current

I = 8 m A

. The radii of inside and outside turns (Fig.) are equal to

a = 50 m m

and

b = 100 m m

Find:
(a) the magnetic induction at the centre of the spiral;
(b) the magnetic moment of the spiral with a given current.

Accepted Answer

(a) From Biot-Savart's law, the magnetic induction due to a circular current carrying wire loop at its centre is given by, B_{r} = \dfrac {\mu_{0}}{2r} i The plane spiral is made up of concentric circular loops, having different radii, varying from  a  to  b . Therefore, the total magnetic induction at the centre, B_{0} = \int \dfrac {\mu_{0}}{2r} dN .... (1) Where  \dfrac {\mu_{0}}{2} i  is the contribution of one turn of radius  r  and  dN  is the number of turns in the interval  (r, r + dr) i.e.  dN = \dfrac {N}{b - a}dr Substituting in equation (1) and integrating the result over  r  between  a  and  b , we obtain, B_{0} = \int_{a}^{b} \dfrac {\mu_{0}i}{2r} \dfrac {N}{(b - a)} dr = \dfrac {\mu_{0} iN}{2(b - a)} ln \dfrac {b}{a} (b) The magnetic moment of a turn of radius  r  is  p_{m} = i\pi r^{2}  and of all turns, p = \int p_{m} dN = \int_{a}^{b} i\pi r^{2} \dfrac {N}{b - a} dr = \dfrac {\pi iN (b^{3} - a^{2})}{3(b - a)} .

Question 8

The maximum charge on the capacitor

Accepted Answer

S_1  and  S_2  are closed for 1 s.Charge on capacitor  q=CE(1-e^{-t/RC})=1-\dfrac{1}{e} ...................(i)The current in inductor , I=\dfrac{E}{R}(1-e^{-tR/L})=1-\dfrac{1}{e} ........................(ii)Now, S_1  and  S_3  are opened and  S_2  is closed,It is LC circuit, q=q_{max}sin(\omega t+\phi) ..........................(iii) I=I_{max} cos (\omega t+\phi) ..................(iv)As total energy (Magnetic +electrical) is constant \dfrac{1}{2}LI^2_{max} \dfrac{1}{2}\dfrac{q^2_max}{C} \dfrac{1}{2}LI^2+\dfrac{1}{2}\dfrac{q^2}{C} ......................................(v) q-{max}=\sqrt{2}(1-\dfrac{1}{e}) ....................................(vi) I_{max} =\sqrt{2}(1-\dfrac{1}{e})  ..............................(vii)From (iii) as t=0. (1-\dfrac{1}{e})=\sqrt{2}(1-\dfrac{1}{e}) sin \phi  \Rightarrow \phi =\dfrac{\pi}{4}  or  \dfrac{3 \pi}{4} After closing  S_2  circuit will be as shown.Direction of current shows that charge on capacitor will be decreasing. Hence \phi= 3 \pi 4 q=\sqrt{2}(1-\dfrac{1}{e}) sin ( \omega t +\dfrac{3 \pi}{4} where,  \omega \dfrac{1}{\sqrt{LC}}=1 q=\sqrt{2}(1-\dfrac{1}{e}) sin (t-\dfrac{3 \pi}{4})

Electromagnetic Induction