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.

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

Torque on current carrying loop in maqgnetic field : Consider a coil PQRS placed in a magnetic field. Let θ be the angle between the plane of the coil and direction of B . When current flow through coil each side experiences a force. The forces on vertical side will constitute a couple. Moment of torque = One of forces × Perpendicular distance between lines of action of force τ = F × QT = F × b sin θ ⇒ τ = IlB b sin θ ⇒ τ = I lb B sin θ ⇒ τ = I A → × B → ( ∵ A = lb ) If there are N turns, then τ = NI A → × B → = M → × B → where M → = NI A → is magnetic moment of loop.

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.

Consider a ring of radius r having n turns per metre. Total number of turns in the windings, N = 2 πrn The magnetic field inside the ring has constant magnitude and tangential direction at every point. ∴ ∮ B → ⋅ d l → = ∮ B d l = B × 2 π r ∴ I e n c l o s e d = N × I = 2 π r n × I = 2 πrnI From Ampere's circuital law, we have ∮ B → ⋅ d l → = μ o μ r I e n c l o s e d B × 2 π r = μ o μ r × 2 π r n I or B = μ 0 μ r n I

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

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

We are given Magnetic moment, M = 6 J / T Magnetic field, B = 0 . 44 T The work done in turning the magnet to align its magnetic moment opposite to the magnetic field, W = M B ( cos θ 1 − cos θ 2 ) Substitute, θ 1 = 60 ° θ 2 = 180 ° We get, W = 6 × 0 . 44 ( cos 60 ° − cos 180 ° ) = 6 × 0 . 44 ( 0 . 5 − - 1 ) = 2 . 64 × 1 . 5 J = 3 . 96 J

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

We are given Magnetic moment, M = 6 J / T Magnetic field, B = 0 . 44 T The work done in turning the magnet to align its magnetic moment normal to the magnetic field, W = M B ( cos θ 1 − cos θ 2 ) Substitute, θ 1 = 60 ° θ 2 = 90 ° We get, W = 6 × 0 . 44 ( cos 60 ° − cos 90 ° ) = 6 × 0 . 44 ( 0 . 5 − 0 ) = 1 . 32 J

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.

Consider that two straight infinite parallel conducting wires with current I 1 and I 2 are separated by a distance x , then force per unit length of each wire is given by : F a = μ 0 I 1 I 2 2 πx where, μ 0 is the magnetic permeability of free space. When the wires carry currents in the opposite direction, they repel and when the wires carry currents in the same direction, they attract. Forces F 3 and F 4 are equal and opposite, so they cancel out each other. Force of repulsion on conductor AB due to EF is F 1 = μ 0 4 π 2 I 1 I 2 a x Force of attraction on conductor CD due to EF is F 2 = μ 0 4 π 2 I 1 I 2 a x + a Net force on the loop is given by F net = F 1 - F 2 F 1 > F 2 ⇒ F net = μ 0 × 2 I 1 I 2 a 4 π 1 x - 1 x + a ⇒ F net = μ 0 4 π 2 I 1 I 2 a 2 x ( x + a )

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

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 4 - Moving Charges and Magnetism>cbse-2020>Q 15

EASY

12th CBSE

IMPORTANT

Earn 100

Define current sensitivity of a galvanometer. Write its expression.

Important Questions on Moving Charges and Magnetism

MEDIUM

12th CBSE

IMPORTANT

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

Derive the expression for the torque acting on the rectangular current carrying coil of a galvanometer. Why is the magnetic field made radial?

MEDIUM

12th CBSE

IMPORTANT

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

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 $= 6.4 \times 10^{- 27} kg$ )

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