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Questions:

Generic 3

9 x 3 = 27

Figure shows the electric field lines around three point charges A, B and C

(i) Which charges are positive?
(ii) Which charge has the largest magnitude? Why?
(iii) In which region or regions of the picture could the electric field be zero? Justify your answer.
(a) Near A
(b) Near B
(c) Near C
(d) Nowhere

Use Lenz’s law to determine the direction of induced current in the situations described by Fig.
(a) A wire of irregular shape turning into a circular shape;
(b) A circular loop being deformed into a narrow straight wire

In a meter bridge, the null point is found at a distance of 40 cm from A. If a resistance of 12\(\Omega \) is connected in parallel with S, the null point occurs at 50 cm from A. Determine the values of R and S.

Show that the force on each plate of a parallel plate capacitor has a magnitude equal to1/2 QE. where Q is the charge on the capacitor and E is the magnitude of electric field between the plates. Explain the origin of the factor 1/2.
Here, we can use the content that the work done in displacing the plates against the force is equal to the increase in energy of the capacitor.

In the figure below the electric field lines on the left have twice the separation of those on the right.
(i) If the magnitude of the field of A is 40 N/C then what force acts on a proton at A?
(ii) What is the magnitude of the field at B?

A 100 turns rectangular coil ABCD (in XY-plane) is hung from one arm of a balance figure. A mass 500 g is added to the other arm to balance the weight of the coil. A current 4.9 A passes through the coil and a constant magnetic field of 0.2 T acting inward (in XZ plane) is switched ON such that only arm CD of length 1 cm lies in the field. How much additional mass m must be added to regain the balance?

The magnetic force applied on CD by magnetic field must balance the weight.

A cell of emf E and Internal resistance r gives a current of 0.5 A with an external resistance of 12 \(\Omega \) and a current of 0.25 A with an external resistance of 25 \(\Omega \) Calculate the
(i) internal resistance of the cell
(ii) emf of the cell

A charge q is placed at the center of a cube of side l. What is the electric passing through each of the cube?

An electric dipole of length 2 cm, when placed with its axis making an angle of 60^o with a uniform electric field, experiences a torque of \(8\sqrt { 3 } \)N-m. Calculate the potential energy of the dipole, if it has a charge of \(\pm \)4nC.

Generic 2

1 x 2 = 2

10.

A charge of 10μC is distributed uniformly over the circumference of a ring of radius 3 m placed on XY-plane with its centre at the origin. Find the electric potential at a point (0,0,4 m).

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Answers Key:

Generic 3

9 x 3 = 27

Electric field lines always start from a positive charge and end at a negative charge. In case of a single charge, electric lines of force start from positive charge and end at infinity.
The magnitude of a charge depends on the number of lines of force emanating from a charge, i.e. higher the number of lines of force, higher the magnitude of charge and vice-versa.
(i) In the given figure, the electric lines of force emanate from A and C. Therefore, charges A and C must be positive.
(ii) The number of electric lines of force emanating is maximum from charge C here, so C must have the largest magnitude.
(iii) Point between two like charges, where electrostatic force is zero, is called neutral point. So, the neutral point lies between A and C only.
Now, the position of neutral point depends on the strength of the forces of charges. Here, more number of electric lines of force show higher strength of charge C than A. So, neutral point lies near A.

(i) Here, the direction of magnetic field is perpendicularly inwards to the plane of paper. If a wire of irregular shape turns into a circular shape, then its area increases (\(\because \) the circular loop has greater area than the loop of irregular shape) so that, the magnetic flux linked also increases. Now, the induced current is produced in a direction such that it decreases the magnetic field [i.e. the current will flow in such a direction, so that the wire forming the loop is pulled inward in all directions (to decrease the area)], i.e current is in anti-clockwise direction, ie. along adcba.
(ii) When a circular loop deforms into a narrow straight wire, the magnetic flux linked with it also decreases. The current induced due to change in flux will flow in such a direction that it will oppose the decrease in magnetic flux, so it will flow anti-clockwise, i.e along adcba due to which the magnetic field produced will be out of the plane of paper.

Applying the condition of balanced Wheatstone bridge,
\(\frac { R }{ S } =\frac { l }{ 100-l } =\frac { 40 }{ 100-40 }\)
\( =\frac { 40 }{ 60 } =\frac { 2 }{ 3 }\)
\(\frac { R }{ S } =\frac { 2 }{ 3 } .........(i)\)
The equivalent resistance of 12\(\Omega \) and S\(\Omega \) in parallel is \(\frac { 12S }{ 12+S } \Omega \)
Again applying the condition \(\frac { R }{ \left( \frac { 12S }{ 12+S } \right) } =\frac { 50 }{ 50 } =1\)
\(\Rightarrow R=\frac { 12S }{ 12+S } \quad ........(iii)\)
From Eqs. (i) and (ii), we get
\(\frac { 2 }{ 3 } S=\frac { 12S }{ 12+S } \)
\(12+S=18 \ or \ S=6\Omega \)
\(R=\frac { 2 }{ 3 } S=\frac { 2 }{ 3 } \times 6=4\Omega\)
\(R=4\Omega \)

Let the distance between the plates be increased by a very small distance\(\triangle x\) = Force x Increased distance
= F.\(\Delta x\) ...(i)

Increase in volume of capacitor
= Area of plates x Increased distance
= A.\(\Delta x\)
u = Energy density = \(\frac { Energy }{ Volume } \)
Increase in energy = u x voulme = u.A.\(\Delta x\) .....(ii)
As, Energy = Work done
F.\(\Delta x\) = u.A.\(\Delta x\) \([\because From \ Eqs.(i) \ and \ (ii)]\)
\(\Rightarrow\) F = u.A
\(=\frac { 1 }{ 2 } \varepsilon _{ 0 }{ E }^{ 2 }.A\quad [\because u=\frac { 1 }{ 2 } \varepsilon _{ 0 }{ E }^{ 2 } \ and \ E=\frac { V }{ d } ]\)
\(=\frac { 1 }{ 2 } \varepsilon _{ 0 }.\frac { { V }^{ 2 } }{ { d }^{ 2 } } .A=\left( \frac { { \varepsilon }_{ 0 }A }{ d } .A \right) \frac { V }{ d } \times \frac { 1 }{ 2 } \)
\(=\frac { 1 }{ 2 } .E.C.V=\frac { 1 }{ 2 } QE \ [\because C=\frac { { \varepsilon }_{ 0 }A }{ d } ,CV=Q]\)

(i) Charge of proton q = 1.6 x 10^-19C
Force on proton at A is F = qE_A
= (1.6 x 10^-19C) (40 N / C)
= 6.4 x 10^-18N
(ii) Since, electric field,
E\(\propto \) Number of electric field lines/Area
Hence, E_B= 1/2 E_A
= 1/2(40 N / c)
= 20 N / C.

For equilibrium balance, net torque should also be equal to zero. When the field is off, \(\Sigma \tau =0\) considering the separation of each hung from mid-point be l.
\( Mgl={ W }_{ coil }l\)
\( \Rightarrow \ 500 \ gl={ W }_{ coil }l\)
\( { W }_{ coil }=500\times9.8N\)
Taking moment of force about mid-point, we have the weight of coil. When the magnetic field is switched ON.
\(Mgl+mgl={ W }_{ coil }l+IBL\sin { { 90 }^{ \circ }l\Rightarrow mgl=BIL \ l } \)
\(\\ m=\frac { BIL }{ g } =\frac { 0.2\times4.91\times1\times{ 10 }^{ -2 } }{ 9.8 } ={ 10 }^{ -3 }kg=1g\)
Thus, 1 g of additional mass must be added to regain the balance.

Let R be external resistance in series with the cell of emf E and internal resistance \(\gamma \). The current in circuit is \(I=\frac { E }{ R+r } \)
Case I: I = 05A, R = 12Ω, then
\(0.5=\frac{E}{12+r}\)
⇒ E = 05 (12 + r)
⇒ E = 6.0 + 05r ......(i)
Casell : I = 0.25A,R = 25Ω ,then
\(0.25=\frac{E}{25+r}\)
⇒ E = 0.25 (25 + r)
⇒ E = 6.25 + 0.25r ......(ii)
From Eqs. (i) and (ii), we get
6.0 + 05 r = 6.25 + 0.25 r
⇒ r = 1Ω
From Eq. (i), we get
E = 6.0 + 05 x (1) = 65V
Hence, (i) internal resistance of the cell is 1 Ω.
(ii) emf of the cell is 6.5 V.

By Gauss' theorem, total electric flux linked with a closed surface is given by
\(\phi =\frac { q }{ { \varepsilon }_{ 0 } } \)
where q is the total charge enclosed by the closed surface
\(\therefore\) Total electric flux linked with cube, \(\phi =\frac { q }{ { \varepsilon }_{ 0 } } \)
As charge is at center, therefore, electric flux is symmetrically distributed through all 6 faces.
Flux linked with each face \(=\frac { 1 }{ 6 } \phi =\frac { 1 }{ 6 } \times \frac { q }{ { \varepsilon }_{ 0 } } =\frac { q }{ 6{ \varepsilon }_{ 0 } } \)

\(Here,length,2a=2cm=2\times { 10 }^{ -2 }m.\)
\(\theta ={ 60 }^{ 0 },\tau =8\sqrt { 3 } N-m\)
Charges, Q = 4\(\times\) 10^-9 C, U = ?
As we know that, \(\tau =Q(2a) E\sin { \theta } \)
\(\Longrightarrow \)Electric field,
\(e=\frac { \tau }{ Q(2a)\sin { \theta } } =\frac { 8\sqrt { 3 } }{ 4\times { 10 }^{ -9 }\times 2\times { 10 }^{ -2 }\times \sin { { 60 }^{ 0 } } } N/C\)
\(\\ \therefore Potential\quad energy, U=-\rho E\cos { \theta } =-Q(2a)E\cos { \theta }\)
\( \\ =-4\times { 10 }^{ -9 }+2\times { 10 }^{ -2 }\times \frac { 8\sqrt { 3 } \times \cos { { 60 }^{ 0 } } }{ 4\times { 10 }^{ -9 }\times 2\times { 10 }^{ -2 }\times \sin { { 60 }^{ 0 } } }\)
\( \\ \Longrightarrow U=-\frac { 8\sqrt { 3 } }{ \sqrt { 3 } } =-8J\)