Describe how a galvanometer can be converted into an ammeter.

A galvanometer is converted into an ammeter by connecting a low resistance in parallel with the galvanometer. This low resistance is called shunt resistance S. The scale is now calibrated in ampere and the range of ammeters depends on the values of the shunt resistance.

A horizontal overhead power line carries a current of 30 A in the east to west direction. What is the magnitude and direction of the magnetic field due to the current at a point 1 . 5 m below the line. μ 0 = 4 π × 10 - 7 T - mA - 1

Given, The current in the power line, I = 90 A . The perpendicular distance of the point from the wire, r = 1 . 5 m . The magnetic filed due to an infinite current carrying wire is given by the relation, B = μ 0 I 2 π r ⇒ B = 4 π × 10 - 7 × 90 2 π × 1 .5 = 1.2 × 10 - 5 T The current is flowing from East to West. The point is below the power line. Hold the wire and place your thumb along the direction of current. Your other fingers will point towards south at the given point. Hence, according to Maxwell's right-hand thumb rule, the direction of the magnetic field at the given point is towards the South.

Use the expression F → = q v → × B → for the force experienced by a charge q moving with a velocity v → in a magnetic field B → to define the SI unit of magnetic field.

Force on a charge (q) moving in a magnetic field B with velocity v → making an angle θ (with the direction of magnetic field B → is given by F m = qvB sin θ = 1 , so or B = F m qv If v = 1 m / s , B = F m qv newton / couloumb . SI unit of magnetic field is tesla. Thus, 1 tesla is the magnetic field in which a charge particle moving with velocity 1 m/s perpendicular to velocity experiences a force of 1 newton/coulomb.

What is a solenoid? Derive the expression for magnetic field due to a long current carrying solenoid by using Ampere's circuital law.

The term solenoid refers to a long, thin loop of wire, often wrapped around a metallic core, which produces a magnetic field when an electric current is passed through it. Solenoids are important because they can create controlled magnetic fields and can be used as electromagnets. Ampere's circuital law states that line integral of magnetic field around any Closed loop is equal to μ ∘ Times the electric current flowing through the cross-section area enclosed by that loop. Mathematically, ∮ B . dl = μI Let the current flowing in the solenoid having number of turns per unit length n be I. Magnitude of magnetic field inside the solenoid is B while at outside is zero. Now ∮ loop B . dl = ∫ B ab . L + ∫ B bc . L ' + ∫ B cd . L + ∫ B de . L ' The value of first term ∫ B ab . L = BL The second and fourth term are zero because angle between magnetic field and the length loop is 90 ° . The third term is also zero as the value of magnetic field outside the solenoid is zero. Total current flowing through the loop I total = nL I From Ampere's circuital law, we get BL = μ ∘ ( nLI ) ⇒ B = μ ∘ nI

What is magnetic dipole moment? Derive the expression for torque on a bar magnet placed in a uniform magnetic field.

The product of the distance (2 l) between the two poles and the pole strength of either pole is called magnetic dipole moment. Its SI unit is 'joule/Tesla' or 'ampere-metre2'. Its direction is from the South Pole towards the North Pole. Torque on a bar magnet placed in a uniform magnetic field Consider a bar magnet NS of length 2l and pole strength m placed in a uniform magnetic field of induction B at an angle θ with the direction of the field (Fig.). Due to the magnetic field B, a force mB acts on the north pole along the direction of the field and a force mB acts on the south pole along the direction opposite to the magnetic field. These two forces are equal and opposite, hence constitute a couple. The torque τ due to the couple is τ = one of the forces ? Perpendicular distance between them τ = F ? NA = mB ? NA . . . ( 1 ) = mB ? 2 l sin θ ∴ τ = MB sin θ . . . ( 2 ) Vectorially Vec τ = Vec M ? Vec B The direction of τ is perpendicular to the plane containing Vec M and Vec B . If B = 1 and θ = 90 ∘ Then from equation ( 2 ) , τ = M Hence, moment of the magnet M is equal to the torque necessary toke the magnet at right angles to a magnetic field of unit magnetic induction.

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 4 - Moving Charges and Magnetism>Manipur Board-2020>Q 8

MEDIUM

12th Manipur Board

IMPORTANT

Earn 100

A current of $\frac{1}{π}$ ampere flows in a circular loop of radius $10 cm$ . Find the magnetic dipole moment of the loop.

Important Questions on Moving Charges and Magnetism

EASY

12th Manipur Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 4 - Moving Charges and Magnetism>Manipur Board-2020>Q 9

In a chamber, a uniform magnetic field of

6.5 \times 10^{6} {ms}^{- 1}

normal to the field. Determine the radius of the circular orbit of the eletron.

(e = 1.6 \times 10^{- 19} C, m_{e} = 9.1 \times 10^{- 31} kg)

MEDIUM

12th Manipur Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 4 - Moving Charges and Magnetism>Manipur Board-2020>Q 10

A circular loop of radius

5 cm

carrying a current of

\frac{10}{π} A

is held in a vertical plane containing north to south direction. The sense of current is clockwise for an observer looking at the coil facing east. Give the magnitude and direction of the magnetic field due to the coil at the centre.