Derive an expression for magnetic potential energy U m = m → . B → , for a magnetic dipole kept in a uniform magnetic field.

The torque on a magnetic dipole of the magnetic moment m → in an external magnetic field B → : τ → = m → × B → In magnitude: τ = m B sin θ .....(i) The magnetic potential energy is given by: U m = ∫ τ θ d θ .....(ii) Using (i) and(ii), we may write: U m = ∫ m B sin θ = - m B cos θ = - m → . B → .....(ii) (ii) is the required expression.

Derive an expression for magnetic potential energy U m = m → . B → , for a magnetic dipole kept in a uniform magnetic field.

The torque on a magnetic dipole of the magnetic moment m → in an external magnetic field B → : τ → = m → × B → In magnitude: τ = m B sin θ .....(i) The magnetic potential energy is given by: U m = ∫ τ θ d θ .....(ii) Using (i) and(ii), we may write: U m = ∫ m B sin θ = - m B cos θ = - m → . B → .....(ii) (ii) is the required expression.

Two rings X and Y are placed in such a way that their axes are along the X and Y axis respectively and their centres are at the origin. Both the Rings X and Y have the same radii of 3 . 14 cm . If the current through X and Y rings are 0 . 3 A and 0 . 4 A respectively then find the value of the resultant magnetic field at the origin. μ ∘ = 4 π × 10 - 7 SI

Given, that the current through X, I x = 0 . 3 A and the current through Y, I y = 0 . 4 A . The magnetic field at the centre can be obtained by using the expression B = μ ∘ I 2 r . Let B x be the magnetic field due to the ring at x-axis, B y be the magnetic field due to the ring at y-axis. Therefore, B x = μ ∘ I x 2 r = 4 π × 10 - 7 × 0 . 3 2 × 3 . 14 = 3 . 76 × 10 - 7 6 . 28 B y = μ ∘ I y 2 r = 4 π × 10 - 7 × 0 . 4 2 × 3 . 14 = 5 . 0265 × 10 - 7 6 . 28 B = B x + B y B = 8 . 78 × 10 - 7 6 . 28 B = 1 . 4 × 10 - 7 T Hence, the magnetic field at the origin is 1 . 4 × 10 - 7 T .

A rectangular coil of 120 turns and an area of 10 × 10 - 4 m 2 is suspended in a radial magnetic field of 45 × 19 - 4 T . If a current of 2 mA through the coil gives it a deflection of 36 ° . Find the effective torsion constant for the spring system holding the coil.

Given the number of turns n = 120 . Area of the coil A = 10 × 10 - 4 m 2 . The radial magnetic field B = 45 × 10 - 4 T . The current in the coil I = 2 mA . Deflection θ = 36 ° ' The current in the galvanometer is given by I = k nBA θ or k = nBAI θ where k = torsion constant of the galvanometer spring, Substituting the values, we have, k = 120 × 45 × 10 - 4 × 10 × 10 - 4 × 2 × 10 - 3 36 = 3 . 0 × 10 - 9 Nm / degree Hence the torsion constant of the galvanometer is 3 × 10 - 9 Nm / degree

Obtained the formula for Lorentz force on a moving electric charge in a uniform electric and magnetic field.

Lorentz for F → = F → m + F → e ce is defined as the combination of the magnetic and electric force on a point charge due to electromagnetic fields. Therefore, the Lorentz forces can be written as : Force due to electric field: When a charged particle carrying a charge q is placed in an electric field , the force experienced by the charge is, F → e = q E → ..........................................(1) This force is directed in the direction of the electric field. Force due to magnetic field: When a charged particle carrying charge q is moving with velocity v → in a magnetic field B → , the force it experiences is given by, F m → = q v → × B → ................................................(2) It is directed perpendicular to the direction of the magnetic field. Thus the total forces is given by, F → = F → e + F → m = q E → + q ( V → × B → ) = q E → + ( v → × B → This is the required equation.

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 4 - Moving Charges and Magnetism>Gujarat Board-2020>Q 17

MEDIUM

12th Gujarat Board

IMPORTANT

Earn 100

Derive an expression for magnetic potential energy $U_{m} = \vec{m} . \vec{B}$ , for a magnetic dipole kept in a uniform magnetic field.

Important Questions on Moving Charges and Magnetism

HARD

12th Gujarat Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 4 - Moving Charges and Magnetism>Gujarat Board-2020>Q 18

Derive an expression for the magnetic field at any point on the axis of a circular current loop.

MEDIUM

12th Gujarat Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 4 - Moving Charges and Magnetism>Gujarat Board-2020>Q 19

A horizontal power line carries a current of $90 A$ in east to west direction. What is the magnitude and direction of the magnetic field due to the current $1.5 m$ below the line? Take $μ_{0} = 4 π \times 10^{- 7} {TmA}^{- 1}$ .