A bar magnet of magnetic moment 6 J / T is aligned at 60 ° with a uniform external magnetic field of 0 · 44 T . Calculate the torque on the magnet when its magnetic moment is aligned opposite to the magnetic field.

We are given Magnetic moment, m = 6 J / T Magnetic field, B = 0 . 44 T Torque on the magnet, τ = m → × B → The torque on the magnet when its aligned opposite to the field, τ = m . B sin θ Substitute, θ = 180 ° τ = 6 × 0 . 44 s i n 180 ° τ = 0

An α ‐ particle and a proton of the same kinetic energy are in turn allowed to pass through a magnetic field B → , acting normal to the direction of motion of the particles. Calculate the ratio of the radii of the circular paths described by them.

Let m α be the mass of the α ‐ particle and m p be the mass of proton. The mass of proton, m p = 1 4 × m α If q p is the charge on proton and q α is the charge on α ‐ particle, then the charge on proton, q p = 1 2 × q α Using the formula, r = mv qB where, r is the radius of the circular path, m is the mass of the particle, q is the charge on the particle in the magnetic field intensity of B . If K is the kinetic energy, then mv = 2 Km where m is the mass and v is the velocity of the particle. So, r = 2 Km qB Since, the kinetic energies are equal, so K , B are constants terms. So, r ∝ m q Let r p , r α be the radii of the circular paths described by the proton and α ‐ particle. ∴ r p r α = m p q p × q α m α ⇒ r p r α = m α 2 × 1 2 × q α × q α m α ⇒ r p r α = 1 1 So, the ratio of the radii of the circular paths described is 1 : 1 .

Explain the significance of a radial magnetic field when a current carrying coil is kept in it.

A magnetic field, in which the plane of the coil in all positions remains parallel to the direction of magnetic field is called radial magnetic field. That is, the angle between the plane of the coil and the magnetic field is zero in all the orientation of the coil. In a radial magnetic field, magnetic torque remains maximum for all the positions of the coil.

A bar magnet of magnetic moment 6 J / T is aligned at 60 ° with a uniform external magnetic field of 0 · 44 T . Calculate the torque on the magnet when its magnetic moment is aligned opposite to the magnetic field.

We are given Magnetic moment, m = 6 J / T Magnetic field, B = 0 . 44 T Torque on the magnet, τ = m → × B → The torque on the magnet when its aligned opposite to the field, τ = m . B sin θ Substitute, θ = 180 ° τ = 6 × 0 . 44 s i n 180 ° τ = 0

Consider a current ( I ) carrying circular coil of radius r and having N turns placed in Y Z plane with its centre at the origin. Derive expression for the value of magnetic field due to it at point ( x , 0 , 0 ) .

The coil is placed in Y Z plane with its centre at the origin. Let P be a point along X - axis at a distance from the origin having coordinates x , 0 , 0 . According to biot-savart law, the magnetic field at a point P due to d l element considered at A is d B → = μ 0 I 4 π d l → × R → R 3 ∵ d l × R = R d l d B = μ 0 I 4 π R d l R 3 = μ 0 I 4 π d l R 2 d B = μ 0 I 4 π d l ( x 2 + r 2 ) A current element opposite to A , that is at B , then the Y - direction component of the magnetic field due to this current element cancel and only X - direction component is only present. So, the net magnetic field along the X - direction. B = ∫ d B x = ∫ d B cos θ = ∫ μ 0 I 4 π d l ( x 2 + r 2 ) cos θ = ∫ μ 0 I 4 π d l ( x 2 + r 2 ) r ( x 2 + r 2 ) 1 2 = μ 0 I 4 π r ( x 2 + r 2 ) 3 2 ∫ d l For a circular coil ∫ d l = 2 πr B = μ 0 I 4 π r ( x 2 + r 2 ) 3 2 × 2 πr B = μ 0 I r 2 2 ( x 2 + r 2 ) 3 2 As the coil having N turns Therefore, B = μ 0 I N r 2 2 ( x 2 + r 2 ) 3 2

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An $α$ -particle is accelerated through a potential difference of $10 kV$ and moves along the x-axis. It enters in a region of the uniform magnetic field $B = 2 \times 10^{- 3} T$ acting the y-axis. Find the radius of its path. (Take mass of $α$ -particle $= 6.4 \times 10^{- 27} kg$ )

Important Questions on Moving Charges and Magnetism

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12th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 4 - Moving Charges and Magnetism>cbse-2018>Q 1

A proton and an electron travelling along parallel paths enter a region of uniform magnetic field, acting perpendicular to their paths. Which of them will move in a circular path with higher frequency?

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12th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 4 - Moving Charges and Magnetism>cbse-2018>Q 2

A bar magnet of magnetic moment

6 J / T

is aligned at

60 °

with a uniform external magnetic field of

0 \cdot 44 T

. Calculate the work done in turning the magnet to align its magnetic moment normal to the magnetic field.

MEDIUM

12th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 4 - Moving Charges and Magnetism>cbse-2018>Q 3

A bar magnet of magnetic moment

6 J / T

is aligned at

60 °

with a uniform external magnetic field of

0 \cdot 44 T

. Calculate the work done in turning the magnet to align its magnetic moment opposite to the magnetic field.

MEDIUM

12th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 4 - Moving Charges and Magnetism>cbse-2018>Q 4

A bar magnet of magnetic moment

6 J / T

is aligned at

60 °

with a uniform external magnetic field of

0 \cdot 44 T

. Calculate the torque on the magnet when its magnetic moment is aligned opposite to the magnetic field.

HARD

12th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 4 - Moving Charges and Magnetism>cbse-2018>Q 5

An iron ring of relative permeability

μ_{r}

has windings of insulated copper wire of

n

turns per metre. When the current in the windings is

I

, find the expression for the magnetic field in the ring.

HARD

12th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 4 - Moving Charges and Magnetism>cbse-2019>Q 6

α ‐

particle and a proton of the same kinetic energy are in turn allowed to pass through a magnetic field

\vec{B}

, acting normal to the direction of motion of the particles. Calculate the ratio of the radii of the circular paths described by them.

HARD

12th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 4 - Moving Charges and Magnetism>cbse-2019>Q 7

A square loop of side $' a'$ carrying a current $I_{2}$ is kept at distance $x$ from an infinitely long straight wire carrying a current $I_{1}$ as shown in the figure. Obtain the expression for the resultant force acting on the loop.

Question Image

HARD

12th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 4 - Moving Charges and Magnetism>cbse-2019>Q 8

Derive the expression for the torque acting on a current carrying loop placed in a magnetic field.

An α-particle is accelerated through a potential difference of 10 kV and moves along the x-axis. It enters in a region of the uniform magnetic field B=2×10-3 T acting the y-axis. Find the radius of its path. (Take mass of α-particle=6.4×10-27 kg)

Important Questions on Moving Charges and Magnetism

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An $α$ -particle is accelerated through a potential difference of $10 kV$ and moves along the x-axis. It enters in a region of the uniform magnetic field $B = 2 \times 10^{- 3} T$ acting the y-axis. Find the radius of its path. (Take mass of $α$ -particle $= 6.4 \times 10^{- 27} kg$ )