The like currents i.e current in both the wire are in the same direction attracts each other. The force is repulsive when the current flows in opposite direction through the wires.
F = \(\frac { { \mu }_{ 0 }{ I }_{ 1 }{ I }_{ 2 }dl }{ 2\pi r } \)
i.e \(F\alpha \frac { 1 }{ r } \)

As the wire of the loop carrying the opposite current is near so the net force acting on the loop is repulsive.

The magnetic field at the center of a circular coil
B = \(\frac { { \mu }_{ 0 }NI }{ 2R } \)
When current I is doubled and radius R is halved,
B' = \(\frac { { \mu }_{ 0 }N\times 2I }{ 2\left( \frac { R }{ 2 } \right) } \)  = 4B
∴ magnetic field becomes four times the original field.

If the electric field is not uniform, then the force experienced by +q is different from that experienced by -q. In addition to the torque, there will be net force acting on the dipole.

R_s = R₁ + R₂ = 15 Ω           (1)
\({ R }_{ p }=\frac { { R }_{ 1 }{ R }_{ 2 } }{ { R }_{ 1 }{ +R }_{ 2 } } =\frac { 56 }{ 15 } \Omega \quad \) (2)
From equation (1) substituting for R₁ + R₂ in equation (2)
\(\frac { { R }_{ 1 }{ R }_{ 2 } }{ 15 } =\frac { 56 }{ 15 } \Omega \)
∴ R₁R₂ = 56
\({ R }_{ 2 }=\frac { 56 }{ 15 } \Omega \)         (3)
Substituting for R₂ in equation (1) from equation (3)
\({ R }_{ 1 }+\frac { 56 }{ { R }_{ 1 } } =15\)
Then, \(\frac { { R }_{ 1 }^{ 2 }+56 }{ { R }_{ 1 } } =15\)
R₁² + 56 = 15 R₁
R₁² - 15 R₁ + 56 = 0
The above equation can be solved using factorisation.
R₁ = 8 Ω (or) R₁ = 7 Ω
If (R₁ = 8 Ω)
Substituting in equation (1)
8 + R₂ = 15
R₂ = 15 – 8 = 7 Ω ,
R₂ = 7 Ω i.e , (when R₁ = 8 Ω ; R₂ = 7 Ω)
If R₁= 7 Ω
Substituting in equation (1)
7 + R₂ = 15
R₂ = 8 Ω , i.e , (when R₁ = 7 Ω ; R₂ = 8 Ω )

If the two spheres are neutral, the angle between them will be 0^o when hanged vertically. Since they are positively charged spheres, there will be a repulsive force between them and they will be at equilibrium with each other at an angle of 30° with the vertical. At equilibrium, each charge experiences zero net force in each direction. We can draw a free-body diagram for one of the charged spheres and apply Newton’s second law for both vertical and horizontal directions.
The free-body diagram is shown below

In the x-direction, the acceleration of the charged sphere is zero.
Using Newton’s second law \((\vec { { F }_{ tot }= } m\vec { a } )\), we have
T sinθ\(\hat { i } \) - F_e\(\hat { i } \) =0
T sinθ = F_e    ......(1)
Here T is the tension acting on the charge due to the string and F_e is the electrostatic force between the two charges.
In the y-direction also, the net acceleration experienced by the charge is zero
Tcosθ\(\hat { j } \) - mg\(\hat { j } \) = 0
Tcosθ = mg   ..(2)
By dividing equation (1) by equation (2),
tanθ = \(\frac { { F }_{ e } }{ mg } \)          .....(3)
Since they are equally charged, the magnitude of the electrostatic force is
\({ F }_{ e }=k\frac { { q }^{ 2 } }{ { r }^{ 2 } } \) where k=\(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \)
Here r = 2a = 2Lsinθ. By substituting these values in equation (3),
tanθ = k\(\frac { { q }^{ 2 } }{ mg(2Lsin\theta )^{ 2 } } \)       ..........(4)
Rearranging the equation (4) to get q
q = 2 Lsinθ\(\\ \sqrt { \frac { mgtan\theta }{ k } } \)
= 2 x 0.1 x sin 30^o x \(\sqrt { \frac { 10^{ -3 }\times 10\times { tan30 }^{ 0 } }{ 9\times 10^{ 9 } } } \)
q = 8.01 x 10^-8C = 80.1 nC

B = 60G,   A = 2 x 10^-4m²
N = 40,     I = 4mA = 4 x 10^-3A
θ = 16⁰
I =\(\frac { K\theta }{ BAN } \)
K = \(\frac { BANI }{ \theta } \)
= \(\frac { 40\times 60\times 2\times 10^{ -4 }\times 4\times 10^{ -3 } }{ 16 } \)
= 1.2 x 10^-4 Nm/degree

Applying kirchhoff's law (loop rule) for loop ABEFA, (since current is reversed, negative sing on both sides)
= R₁ x 0 - 3 x I₁ - 2I₁ = -6-3-1
= 3I₁ - 2I₁= -10
= 5I₁ = -10
I₁ = 2A
for loop BCDEB
= R x I₁ - R₁ x 0 = -4 + 3
= I₁R = 1
\(R=\cfrac { 1 }{ 2 } \Omega \)
Potential difference between A and D through path ABCD is
= 6-4 + V_AD = I₁R
= 10 + V_AD = -2 x \(\cfrac { 1 }{ 2 } \)
= 10 + CAD = -1
V_AD = -9 Volt

According to the superposition principle, the total electric potential at a point is equal to the sum of the potentials due to each charge at that point.
Consider the point at which the total potential zero is located at a distance x from the charge +q as shown in the figure.

The total electric potential at P is zero.
V_tot = \(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \left( \frac { q }{ x } -\frac { 2q }{ (9-x) } \right) \)=0
Which gives \(\frac { q }{ x } -\frac { 2q }{ (9-x) } \)
or \(\frac { 1 }{ x } =\frac { 2 }{ (9-x) } \)
Hence, x = 3m

The electric flux of D is less than that of C
The electric flux of C is less than that of B
The electric flux of B is less than that of A

Yellow - 4
Violet - 7
Orange - 10³
Silver - Tolerance - 10%

The bridge wire is soldered at the ends of the copper strips. Due to imperfect contact, some resistance, might be introduced at the contact. These are called end resistances. This error can be eliminated, if another set of readings are taken with P and Q interchanged and the average value of P is found.

The substances which have an abundance of free electrons are called conductors. These free electrons move at random throughout the conductor at a given temperature. In general due to this random motion, there is no net transfer of charges from one end of the conductor to other end and hence no current. When a potential difference is applied by the battery across the ends of the conductor, the free electrons drift towards the positive terminal of the battery, producing a net electric current.

(i) In polar molecules, the centers of the positive and negative charges are separated even in the absence of an external electric field.
(ii) They have a permanent dipole moment.
(iii) Examples : H₂O, N₂O, HCI and NH₃.

Electric field lines are a set of continuous lines which represent the electric field in some region of space visually.

The internal resistance of a cell is the resistance offered to the flow of current (by the electrolyte) inside the cell.