Q.23 (i) The electric field at a point on the equatorial line of an electric dipole can be derived as follows:

Step 1: Consider an electric dipole with charges +q and -q separated by a distance 2a. Let P be a point on the equatorial line of the dipole at a distance r from the center O of the dipole.

Step 2: The electric field at P due to the positive charge +q is E1 = kq/r² in the direction OP.

Step 3: The electric field at P due to the negative charge -q is E2 = kq/(r+2a)² in the direction PO.

Step 4: Since E1 and E2 are in opposite directions, the net electric field at P is E = E1 - E2 = kq/r² - kq/(r+2a)².

(ii) The orientation of the dipole in a uniform electric field is as follows:

(a) Stable equilibrium: The dipole aligns itself along the direction of the electric field. The negative charge is closer to the direction from which the field lines are coming and the positive charge is in the direction in which the field lines are going.

(b) Unstable equilibrium: The dipole aligns itself opposite to the direction of the electric field. The positive charge is closer to the direction from which the field lines are coming and the negative charge is in the direction in which the field lines are going.

Q.24 The equation of the balanced state in a Wheatstone bridge using Kirchhoff's laws can be derived as follows:

Step 1: According to Kirchhoff's first law, the sum of currents entering a junction is equal to the sum of currents leaving the junction.

Step 2: According to Kirchhoff's second law, the sum of the electromotive forces in any closed loop or mesh in a network is equal to the sum of the potential drops in that loop.

Step 3: In a balanced Wheatstone bridge, no current flows through the galvanometer. Therefore, the potential difference between the two junctions connected by the galvanometer is zero.

Step 4: From these, we can derive the condition for a balanced Wheatstone bridge: (R1/R2) = (R3/R4), where R1, R2, R3, and R4 are the resistances in the four arms of the bridge.

Q.25 The magnetic field lines for a current carrying solenoid when a rod made of (i) copper, (ii) aluminium and (iii) iron are inserted within the solenoid are as follows:

(i) Copper: The magnetic field lines will be uniform and parallel to each other within the solenoid. Copper being a non-magnetic material does not affect the magnetic field.

(ii) Aluminium: Similar to copper, aluminium being a non-magnetic material does not affect the magnetic field. The field lines remain uniform and parallel within the solenoid.

(iii) Iron: Iron being a ferromagnetic material, it enhances the magnetic field within the solenoid. The field lines are more concentrated within the iron rod.

Question

Q.23 (i) The electric field at a point on the equatorial line of an electric dipole can be derived as follows:

Step 1: Consider an electric dipole with charges +q and -q separated by a distance 2a. Let P be a point on the equatorial line of the dipole at a distance r from the center O of the dipole.

Step 2: The electric field at P due to the positive charge +q is E1 = kq/r² in the direction OP.

Step 3: The electric field at P due to the negative charge -q is E2 = kq/(r+2a)² in the direction PO.

Step 4: Since E1 and E2 are in opposite directions, the net electric field at P is E = E1 - E2 = kq/r² - kq/(r+2a)².

(ii) The orientation of the dipole in a uniform electric field is as follows:

(a) Stable equilibrium: The dipole aligns itself along the direction of the electric field. The negative charge is closer to the direction from which the field lines are coming and the positive charge is in the direction in which the field lines are going.

(b) Unstable equilibrium: The dipole aligns itself opposite to the direction of the electric field. The positive charge is closer to the direction from which the field lines are coming and the negative charge is in the direction in which the field lines are going.

Q.24 The equation of the balanced state in a Wheatstone bridge using Kirchhoff's laws can be derived as follows:

Step 1: According to Kirchhoff's first law, the sum of currents entering a junction is equal to the sum of currents leaving the junction.

Step 2: According to Kirchhoff's second law, the sum of the electromotive forces in any closed loop or mesh in a network is equal to the sum of the potential drops in that loop.

Step 3: In a balanced Wheatstone bridge, no current flows through the galvanometer. Therefore, the potential difference between the two junctions connected by the galvanometer is zero.

Step 4: From these, we can derive the condition for a balanced Wheatstone bridge: (R1/R2) = (R3/R4), where R1, R2, R3, and R4 are the resistances in the four arms of the bridge.

Q.25 The magnetic field lines for a current carrying solenoid when a rod made of (i) copper, (ii) aluminium and (iii) iron are inserted within the solenoid are as follows:

(i) Copper: The magnetic field lines will be uniform and parallel to each other within the solenoid. Copper being a non-magnetic material does not affect the magnetic field.

(ii) Aluminium: Similar to copper, aluminium being a non-magnetic material does not affect the magnetic field. The field lines remain uniform and parallel within the solenoid.

(iii) Iron: Iron being a ferromagnetic material, it enhances the magnetic field within the solenoid. The field lines are more concentrated within the iron rod.

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