## A charged ring of mass m = 50 gm, charge 2 coulomb and radius R = 2 m is placed on a smooth horizontal surface….

Q: A charged ring of mass m = 50 gm, charge 2 coulomb and radius R = 2 m is placed on a smooth horizontal surface. A magnetic field varying with time at a rate of (0.2 t) Tesla/sec is applied on to the ring in a direction normal to the surface of ring. Find the angular speed attained in a time t1 = 10 sec.

Solution :Induced EMF is $\displaystyle \oint \vec{E}.\vec{dl} = \frac{d \phi}{dt}$

$\displaystyle E (2 \pi R) = \pi R^2 \frac{d B}{dt}$

$\displaystyle E = \frac{R}{2} \times 0.2 t$

$\displaystyle E = \frac{2}{2} \times 0.2 t = 0.2 t$

As Force F = q E = 0.4 t

Torque τ = I α = F R

α = F R/I

$\displaystyle \alpha = \frac{F R}{m R^2}$

$\displaystyle \alpha = \frac{F}{m R} = \frac{0.4 t}{50 \times 10^{-3} \times 2}$

α = 4 t

$\displaystyle \frac{d\omega}{dt} = 4 t$

$\displaystyle \omega = \int_{0}^{10}4 t dt$

$\displaystyle \omega = [\frac{4t^2}{2}]_{0}^{10}$

## There exists a uniform cylindrically symmetric magnetic field directed along the axis of a cylinder …

Q: There exists a uniform cylindrically symmetric magnetic field directed along the axis of a cylinder but varying with time as B = k t. If an electron is released from rest in this field at a distance of ‘ r ‘ from the axis of cylinder, its acceleration, just after it is released would be (e and m are the electronic charge and mass respectively)

Solution :$\displaystyle \oint \vec{E}.\vec{dl} = \frac{d \phi}{dt}$

$\displaystyle E (2 \pi r) = \pi r^2 \frac{d B}{dt}$

$\displaystyle E = \frac{r}{2} \frac{d B}{dt}$

$\displaystyle E = \frac{r}{2} \frac{d}{dt}(k t)$

$\displaystyle E = \frac{r}{2} k$

Acceleration $\displaystyle a = \frac{e E}{m}$

$\displaystyle a = \frac{e}{m} \frac{k r}{2}$

$\displaystyle a = \frac{e r k}{2 m}$

Directed along tangent to the circle of radius r whose center lies on the axis of cylinder.

## A conducting circular loop is placed in a uniform magnetic field of 0.02 T, with its plane perpendicular to the field….

Q: A conducting circular loop is placed in a uniform magnetic field of 0.02 T, with its plane perpendicular to the field. If the radius of the loop starts shrinking at a constant rate of 1.0 mm/s, then find the emf induced in the loop, at the instant when the radius is 4 cm.

Solution : Magnetic flux linked with the loop is φ = B . π r2

Induced emf $\displaystyle e = \frac{d\phi}{dt}$

$\displaystyle e = \frac{d}{dt}(B . \pi r^2)$

$\displaystyle e = B . \pi . 2r \frac{dr}{dt}$

$\displaystyle e = 0.02 \times 3.14 \times 2 \times 4 \times 10^{-2} \times 1 \times 10^{-3}$

## A uniform magnetic field of 0.08 T is directed into the plane of the page and perpendicular to it as shown in the figure….

Q: A uniform magnetic field of 0.08 T is directed into the plane of the page and perpendicular to it as shown in the figure. A wire loop in the plane of the page has constant area 0.010 m2 . The magnitude of magnetic field decrease at a constant rate of 3.0 × 10–4 Ts–1 . Find the magnitude and direction of the induced emf in the loop.

Solution : Magnitude of Induced emf is

$\displaystyle |e| = \frac{d\phi}{dt}$

$\displaystyle |e| = \frac{d}{dt}(B A)$

$\displaystyle |e| = A \frac{d B}{dt}$

$\displaystyle |e| = 0.010 \times 3 \times 10^{-4} volt$

e = 3 × 10-6 volt

Direction is Clockwise according to Lenz’s Law

## The horizontal component of the earth’s magnetic field at a place is 3 × 10^–4 T and the dip is tan–1(4/3)…..

Q: The horizontal component of the earth’s magnetic field at a place is 3 × 10–4 T and the dip is tan–1(4/3). A metal rod of length 0.25 m placed in the north-south position is moved at a constant speed of 10 cm/s towards the east. Find the e.m.f induced in the rod .

Solution : Angle of dip = tan–1(4/3)

Angle of dip is $\displaystyle tan \delta = \frac{B_V}{B_H}$

$\displaystyle \frac{4}{3} = \frac{B_V}{B_H}$

$\displaystyle \frac{4}{3} = \frac{B_V}{3 \times 10^{-4}}$

$\displaystyle B_V = \frac{4}{3} \times ( 3 \times 10^{-4})$

$\displaystyle B_V = 4 \times 10^{-4}$

Induced EMF = BV . l . v

$\displaystyle e = 4 \times 10^{-4} \times 0.25 \times 0.10$

= 10 × 10–6 volt

## Two long parallel conducting horizontal rails are connected by a conducting wire at one end. A uniform magnetic field B exists in the region ….

Q: Two long parallel conducting horizontal rails are connected by a conducting wire at one end. A uniform magnetic field B exists in the region of space. A light uniform ring of diameter d which is practically equal to separation between the rails, is
placed over the rails as shown in the figure. If resistance of ring is λ per unit length, calculate the force required to pull the ring with uniform velocity v.

Solution : Emf induced in the conducting wire is

$\displaystyle e = B (2R) v = B d v$

Resistance $r = \frac{\lambda \pi d}{2}$

Current $\displaystyle I = \frac{e}{r}$

$\displaystyle I = \frac{B v d}{\frac{\lambda \pi d}{2}}$

$\displaystyle I = \frac{2 B v}{\pi \lambda}$

Force required to pull the ring is , F = 2 (Retarding force on each semicircle )

F = I d B + I d B = 2 I d B

$\displaystyle F = 2 (\frac{2 B v}{\pi \lambda}) d B$

$\displaystyle F = (\frac{4 B^2 d v}{\pi \lambda})$

## A horizontal wire is free to slide on the vertical rails of a conducting frame as shown in figure…

Q: A horizontal wire is free to slide on the vertical rails of a conducting frame as shown in figure. The wire has a mass m and length l and the resistance of the circuit is R. If a uniform magnetic field B is directed perpendicular to the frame, then find the terminal speed of the wire as it falls under the force of gravity .

Solution : Emf induced in the wire moving in the magnetic field = B l v

Induced current $\displaystyle I = \frac{e}{R} = \frac{B l v}{R}$

Magnetic Force acting on the wire is $\displaystyle F_m = I l B$

$\displaystyle F_m = (\frac{B l v}{R}) l B = \frac{B^2 l^2 v}{R}$

At terminal Speed ,

$\displaystyle m g = F_m$

$\displaystyle m g = \frac{B^2 l^2 v}{R}$

$\displaystyle v = \frac{m g R}{B^2 l^2 }$

## A long straight wire is arranged along the symmetry axis of a toroidal coil of rectangular cross-section, whose dimensions are given in the figure…

Q: A long straight wire is arranged along the symmetry axis of a toroidal coil of rectangular cross-section, whose dimensions are given in the figure. The number of turns on the coil is N, and relative permeability of the surrounding medium is unity. Find the amplitude of the emf induced in this coil, if the current I = I0 cos ωt flows along the straight wire.

Solution : Magnetic field at a distance  x from long straight wire is

$\displaystyle B = \frac{\mu_0 N I}{2 \pi x}$

Magnetic flux linked with elemental area dA is

dφ = B . dA = B (h dx)

$\displaystyle d\phi = \frac{\mu_0 N I}{2 \pi x} (h dx)$

Total magnetic flux $\displaystyle \phi = \int_{a}^{b} \frac{\mu_0 N I}{2 \pi x} (h dx)$

$\displaystyle \phi = \frac{\mu_0 N I h}{2 \pi } \int_{a}^{b} \frac{dx}{x}$

$\displaystyle \phi = \frac{\mu_0 N I h}{2 \pi } [ln x]_{a}^{b}$

$\displaystyle \phi = \frac{\mu_0 N I h}{2 \pi } [ln a – ln b]$

$\displaystyle \phi = \frac{\mu_0 N I h}{2 \pi } ln\frac{a}{b}$

$\displaystyle \phi = \frac{\mu_0 N h (I_0 cos\omega t )}{2 \pi } ln\frac{a}{b}$

$\displaystyle e = – \frac{d \phi}{dt} = – \frac{\mu_0 N h }{2 \pi } ln \frac{a}{b} \frac{d}{dt}(I_0 cos\omega t)$

$\displaystyle e = -\frac{\mu_0 N h }{2 \pi } ln \frac{a}{b} (-I_0 sin\omega t) \omega$

$\displaystyle e = \frac{\mu_0 N h \omega }{2 \pi } (I_0 sin\omega t) ln \frac{a}{b}$

## A rectangular loop with current I has dimension as shown in figure. Find the magnetic flux φ through the infinite region …

Q: A rectangular loop with current I has dimension as shown in figure. Find the magnetic flux φ through the infinite region to the right of line PQ.

solution : Magnetic field at a distance due to long wire is

$\displaystyle B = \frac{\mu_0}{4\pi}\frac{2I}{x}$

Magnetic flux linked with small area is , dφ = B . dA

$\displaystyle d\phi = \frac{\mu_0}{4\pi}\frac{2I}{x} (l dx)$

Total magnetic flux

$\displaystyle \phi = \frac{\mu_0 I l}{2\pi}\int_{a}^{a+b} \frac{dx}{x}$

$\displaystyle \phi = \frac{\mu_0 I l}{2\pi} [ln x]_{a}^{a+b}$

$\displaystyle \phi = \frac{\mu_0 I l}{2\pi} [ln (a+b) – ln a]$

$\displaystyle \phi = \frac{\mu_0 I l}{2\pi} ln (\frac{a+b}{a})$

$\displaystyle \phi = \frac{\mu_0 I l}{2\pi} ln (1 + \frac{b}{a})$

## An emf of 15 volt is applied in a circuit containing 5 H inductance and 10 Ω resistance. Find the ratio of ….

Q: An emf of 15 volt is applied in a circuit containing 5 H inductance and 10 Ω resistance. Find the ratio of the currents at time t = ∞ and t = 1 second.

Solution : $\displaystyle I = I_0 (1 -e^{- \frac{Rt}{L}})$

At t = ∞ ; I = I0

I0 =15/10 = 3/2 A

At t = 1 sec

$\displaystyle I_1 = I_0 (1 -e^{-2})$

$\displaystyle \frac{I_1}{I_0} = (1 – \frac{1}{e^2})$

$\displaystyle \frac{I_1}{I_0} = \frac{e^2 -1}{e^2}$