CBSE Class 12 Physics Magnetic Effects of Electric Current Assignment

CBSE Class 12 Physics Magnetic Effects of Electric Current Assignment. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

Comprehension-1

Magnetic field intensity (B) due to current carrying conductor can be calculated by use of Biot-savart law. which is dB =μ₀/4∏ Idlxr/r³, where dB is magnetic field due current element Idl at a position r from current element. For straight wire carrying current magneticfield at a distance R from wire isB=μ₀/4∏ I/R(sin α + sin β) and magnetic field due to a circular arc at its centre is B =μ₀I/4∏R. θ where θ angle of circular arc at centre, R is radius of circular arc.

Now answer the following questions.
1. The magnetic field at C due to curved part is

2. A wire loop carrying a current I is shown in figure. The magnetic field induction at C due to straight part is

3. The net magnetic field at C due to the current carrying loop is

Comprehension-2

A current carrying coil behave like short magnet whose magnetic dipole moment M = n I A. Where direction of M is taken along the direction of magnetic field on its axis and n is no of turms A is area of coil and I is current flowing through coil. When such a coil is put in magnetic field (B) magnetic torque (τ) acts over it as τ = M × B and potential energy in of the current loop in the magnetic field is u = – M · B.

4. A current of 3 A is flowing in a plane circular coil of radius 4 cm and having 20 turns. The coil is placed in a uniform magnetic field of 0.5 Wb m–2. Then, the dipole moment of the coil is
(a) 3000 A m² (b) 0.3 A m² (c) 75 A m² (d) 1.88 × 10^–2 A m²

5. A current of 3 A is flowing in a plane circular coil of radius 4 cm and having 20 turns. The coil is placed in a uniform magnetic field of 0.5 Wb m^–2. Then, the P.E. of the magnetic dipole in the position of stable equilibrium is
(a) –1500 J (b) –9.4 mJ (c) +0.15 J (d) +1500 J

6. In above question, to hold the current-carrying coil with normal to its plane making an angle of 90° with the direction of magnetic induction, the necessary torque is
(a) 1500 Nm (b) 9.4 × 10^–3 Nm (c) 15 Nm (d) 150 Nm

Each of the questions given below consists of two statements, an assertion (A) and reason (R).
Select the number corresponding to the appropriate alternative as follows
(a) If both A and R are true and R is the correct explanation of A
(b) If both A and R are true but R is not the correct explanation of A
(c) If A is true but R is false
(d) If A is false but R is true

7. A: The magnetic field interacts with a moving charge and not with a stationary charge.
R: A moving charge is a source of magnetic field.

8. A: When a charged particle moves in a magnetic field, its kinetic energy remains the same.
R: Force on a moving charged particle due to magnetic field is always perpendicular to velocity vector

9. A: The two charged particle of same mass and charge are describing circle in the same magnetic field with radii r₁ and r₂ (>r₁), the speed of first particle is less than that of the second particle.
R: v = qBr/m i.e. v ∝ r, so the speed of the first particle is less than that of the second particle
( r₂ > r₁)

10. A: A current is flowing north along a power line. The direction of magnetic field above it, neglecting the earth’s field is towards east.
R: Apply Flemming’s left hand rule.

11. A: A charge can go undeviated in the combined effect of electric and magnetic field.
R: If v || E, then electric force does not change the direction of motion and when v || B, the magnetic force on the test charge is zero.

12. A: When a charged particle is fired in a magnetic field, the centripetal force on it is independent of mass of the particle.
R: The centripetal force on a test charge q₀ moving with velocity v in a magnetic field B is F_m = q₀v × B

13. A: Angle of dip at the equator is zero.
R: At the equator, the vertical component of earth’s magnetic field is zero.

14. A: Magnetic susceptibility has no units and no dimensions.
R: Xm = H/I, where I is the intensity of magnetisation and H is the magnetising field.

Miscellaneous Assignment
1. (a) 2. (a) 3. (d) 4. (d) 5. (b) 6. (b) 7. (a) 8. (a) 9. (a) 10. (a) 11. (b) 12. (a) 13. (a) 14. (a)