Rajik

Ashok0210

Asked: February 24, 2021In: Class 12 Physics

If the bar magnet in exercise 5.13 is turned around by 180º, where will the new null points be located?

Class 12 Physics CBSE and UP Board Magnetism and Matter Chapter-5 Exercise 5.14 NCERT Solutions for Class 12 Physics Chapter 5 Question 14

Ashok0210

Asked: February 24, 2021In: Class 12 Physics

A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at 14 cm from the centre of the magnet. The earth’s magnetic field at the place is 0.36 G and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null–point (i.e., 14 cm) from the centre of the magnet? (At null points, field due to a magnet is equal and opposite to the horizontal component of earth’s magnetic field.)

Class 12 Physics CBSE and UP Board Magnetism and Matter Chapter-5 Exercise 5.13 NCERT Solutions for Class 12 Physics Chapter 5 Question 13

Ashok0210

Asked: February 24, 2021In: Class 12 Physics

A short bar magnet has a magnetic moment of 0.48 J T⁻¹. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on (a) the axis, (b) the equatorial lines (normal bisector) of the magnet.

Class 12 Physics CBSE and UP Board Magnetism and Matter Chapter-5 Exercise 5.12 NCERT Solutions for Class 12 Physics Chapter 5 Question 12

Ashok0210

Asked: February 24, 2021In: Class 12 Physics

At a certain location in Africa, a compass points 12º west of the geographic north. The north tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian points 60º above the horizontal. The horizontal component of the earth’s field is measured to be 0.16 G. Specify the direction and magnitude of the earth’s field at the location.

Class 12 Physics CBSE and UP Board Magnetism and Matter Chapter-5 Exercise 5.11 NCERT Solutions for Class 12 Physics Chapter 5 Question 11

Ashok0210

Asked: February 24, 2021In: Class 12 Physics

A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 22º with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.35 G. Determine the magnitude of the earth’s magnetic field at the place.

Class 12 Physics CBSE and UP Board Magnetism and Matter Chapter-5 Exercise 5.10 NCERT Solutions for Class 12 Physics Chapter 5 Question 10

Ashok0210

Asked: February 24, 2021In: Class 12 Physics

A circular coil of 16 turns and radius 10 cm carrying a current of 0.75 A rests with its plane normal to an external field of magnitude 5.0 × 10⁻² T. The coil is free to turn about an axis in its plane perpendicular to the field direction. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a frequency of 2.0 s⁻¹. What is the moment of inertia of the coil about its axis of rotation?

Class 12 Physics CBSE and UP Board Magnetism and Matter Chapter-5 Exercise 5.9 NCERT Solutions for Class 12 Physics Chapter 5 Question 9

Ashok0210

Asked: February 24, 2021In: Class 12 Physics

A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 10⁻⁴ m² , carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane. (a) What is the magnetic moment associated with the solenoid? (b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of 7.5 × 10⁻² T is set up at an angle of 30º with the axis of the solenoid?

Class 12 Physics CBSE and UP Board Magnetism and Matter Chapter-5 Exercise 5.8 NCERT Solutions for Class 12 Physics Chapter 5 Question 8

Ashok0210

Asked: February 24, 2021In: Class 12 Physics

A bar magnet of magnetic moment 1.5 J T⁻¹ lies aligned with the direction of a uniform magnetic field of 0.22 T. (a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: (i) normal to the field direction, (ii) opposite to the field direction? (b) What is the torque on the magnet in cases (i) and (ii)?

Class 12 Physics CBSE and UP Board Magnetism and Matter Chapter-5 Exercise 5.7 NCERT Solutions for Class 12 Physics Chapter 5 Question 7

Ashok0210

Asked: February 24, 2021In: Class 12 Physics

If the solenoid in Exercise 5.5 is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of 30° with the direction of applied field?

Class 12 Physics CBSE and UP Board Magnetism and Matter Chapter-5 Exercise 5.6 NCERT Solutions for Class 12 Physics Chapter 5 Question 6

Rajik

If the bar magnet in exercise 5.13 is turned around by 180º, where will the new null points be located?

A short bar magnet has a magnetic moment of 0.48 J T⁻¹. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on (a) the axis, (b) the equatorial lines (normal bisector) of the magnet.

If the solenoid in Exercise 5.5 is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of 30° with the direction of applied field?

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