A straight wire of mass 200 g and length 1.5 m carries a current of 2 A. It is suspended in mid-air by a uniform horizontal magnetic field B. What is the magnitude of the magnetic field?
If the magnetic field is parallel to the positive y-axis and the charged particle is moving along the positive x-axis, which way would the Lorentz force be for (a) an electron (negative charge), (b) a proton (positive charge).
An element ∆x iˆ is placed at the origin and carries a large current i = 10 A. What is the magnetic field on the y-axis at a distance of 0.5 m? ∆x = 1 cm. Answer: 4 × 10–8 T.
The direction of the field is in the +z-direction.
1. state right-hand thumb rule.
answer: Curl the palm of your right hand around the circular wire with the fingers pointing in the direction of the current. The right-hand thumb gives the direction of the magnetic field.
2. Write mathematical expression for magnetic field at the centre of a current carrying circular coil.
3. On what factors the magnetic field at the centre of a current carrying coil depends?
4. Name and state the rule for finding the directions of a magnetic field due to a circular current loop.
5. Name the physical quantity whose SI unit is wb/m2
4.1 A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?
6. A circular coil of radius ‘R’ carries a current ‘I’. Write the expression for the
magnetic field due to this coil at its centre. Find out the direction of the magnetic field.
1. State ampere's circuital Law. Give its mathematical expression.
2. State Ampere’s Circuital Law. Derive an expression for the magnetic field at a
point due to straight current carrying conductor.
3. An electric current is flowing due south along a power line. What is the direction of the magnetic field at a point above it? name the rule used to find it.
4.2 A long straight wire carries a current of 35 A. What is the magnitude of the field B at a point 20 cm from the wire?
4.3 A long straight wire in the horizontal plane carries a current of 50 A in north to south direction. Give the magnitude and direction of B at a point 2.5 m east of the wire.
4.4 A horizontal overhead 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?
5. Find the value of magnetic field inside a hollow straight current carrying conductor at a distance r from axis of the loop. Ans B=0
1. Derive an expression for the magnetic field at a point along the axis of an air cored long solenoid, using a Ampere’s circuital law.
2. A solenoid is 1m long and 3 cm in mean diameter. It has 5 layers of windings of 800 turns each and carries a current of 5 A. Find Magnetic Field Induction at the center of the solenoid.
Ans: 2.5 x 10^-2 T, parallel to the axis of the solenoid.
1. State the rule which is used to find the direction of force on a charge moving in a per pendicular magnetic field.
2. What is meant by Lorentz Force? Give its mathematical expression.
3. Write the expression for force acting on any charge q and velocity v that enters in
magnetic field in vector form.
4. Name the physical quantity whose SI unit is wb/m2
5. Under what condition does an electron moving through a magnetic field experience
maximum force?
6. Under what condition is a force acting on a charge moving through a uniform magnetic field minimum?
7.What is the direction of the force acting on a charged particle Q moving with a velocity V in a uniform magnetic field B?
8. What will be the path of a charged particle moving perpendicular to a uniform magnetic field?
Among alpha, beta and gamma radiations which get deflected by a magnetic field?
9. Electron being projected along the positive X- axis experience is a force due to magnetic field along the Y-axis. What is the direction of magnetic field?
10. A beam of Alpha particles projected along the positive X- axis experiences a force due to a magnetic field along the positive y-axis. What is the direction of the magnetic field?
11. Which of the following will experience maximum force when projected with the same velocity v perpendicular to the magnetic field (i) Alpha particle or (ii) beta particle?
12. An electron does not suffer any deflection while passing through a region of uniform magnetic field. What is the direction of the magnetic field?
13. What amount of work is done by a magnetic force in a moving charge and why?
1) Derive mathematical expression for the magnitude of the magnetic force per unit of length between two parallel straight current carrying conductors. Hence define 1ampere electric current.
2) Two parallel wires carry equal currents in same direction. Write value of magnetic field due to the two wires at a centre point between wires.
3) The wires which connect the battery of an automobile to its starting motor carry a current of 300 A (for a short time). What is the force per unit length between the wires if they are 70 cm long and 1.5 cm apart? Is the force attractive or repulsive?
1. A compass needle free to turn in a horizontal plane is placed at the centre of circular coil of 30 turns and radius 12 cm. The coil is in a vertical plane making an angle of 45º with the magnetic meridian. When the current in the coil is 0.35 A, the needle points west to east.
(a) Determine the horizontal component of the earth’s magnetic field at the location.
(b) The current in the coil is reversed, and the coil is rotated about its vertical axis by an angle of 90º in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the places to be zero.
1) Derive mathematical expression for the magnitude of the magnetic force per unit of length between two parallel straight current carrying conductors. Hence define 1ampere electric current.
2) Two parallel wires carry equal currents in same direction. Write value of magnetic field due to the two wires at a centre point between wires.
3) The wires which connect the battery of an automobile to its starting motor carry a current of 300 A (for a short time). What is the force per unit length between the wires if they are 70 cm long and 1.5 cm apart? Is the force attractive or repulsive?
1. A compass needle free to turn in a horizontal plane is placed at the centre of circular coil of 30 turns and radius 12 cm. The coil is in a vertical plane making an angle of 45º with the magnetic meridian. When the current in the coil is 0.35 A, the needle points west to east.
(a) Determine the horizontal component of the earth’s magnetic field at the location.
(b) The current in the coil is reversed, and the coil is rotated about its vertical axis by an angle of 90º in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the places to be zero.
1.Derive an expression for torque acting on a rectangular current carrying loop kept in a uniform magnetic field B. Indicate the direction of torque acting on the loop.
2. what is the value of net force on a plane current loop kept in uniform magnetic field?
3. Does the value of torque depend on the shape of the plane coil?
4. A circular coil of 20 turns and radius 10 cm is placed in a uniform magnetic field of 0.10 T normal to the plane of the coil. If the current in the coil is 5.0 A, what is the
(a) total torque on the coil,
(b) total force on the coil,
(c) average force on each electron in the coil due to the magnetic field?
(The coil is made of copper wire of cross-sectional area 10–5 m2, and the free electron density in copper is given to be about 1029 m–3.)
5. A uniform magnetic field of 3000 G is established along the positive z-direction. A rectangular loop of sides 10 cm and 5 cm carries a current of 12 A. What is the torque on the loop in the different cases shown in Figure? What is the force on each case? Which case corresponds to stable equilibrium?
(k/BNA) is called galvanometer constant.
Qn. Define current sensitivity and voltage sensitivity of a galvanometer.
Deflection produced by unit current in the galvanometer is called its current sensitivity. θ/i = (BNA/k)
The deflection produced by a unit potential difference across the galvanometer coil is called voltage sensitivity. θ/iR = (BNA/kR)
R = resistance of the galvanometer coil.
Qn. How will you convert a galvanometer into a voltmeter?
A very large valued resistance Rs is connected in series with the galvanometer coil.
Total resistance of the voltmeter Rv = Rs+R
Current for maximum deflection is ig. The series resistance R can be calculated using ohm's law V=ig(R+Rs)
Qn. A galvanometer coil has a resistance of 12 Ω and the metre shows full scale deflection for a current of 3 mA. How will you convert the metre into a voltmeter of range 0 to 18 V?
Qn. How will you convert a galvanometer into an ammeter?
Answer: A very small valued resistance Rs called shunt is connected in parallel to the galvanometer coil.
The shunt resistance Rs can be calculated using ohm's law
Rs.(i-ig)=R.ig
Total resistance Ra of ammeter is calculated as 1/Ra = 1/Rs +1/R
Qn. A galvanometer coil has a resistance of 15 Ω and the metre shows full scale deflection for a current of 4 mA. How will you convert the metre into an ammeter of range 0 to 6 A?
Qn. Find expression for the torque on a rectangular current loop in a uniform magnetic field
Answer: A rectangular loop carrying a steady current i and placed in a uniform magnetic field experiences a torque.
Let the plane of the loop makes angle α with the magnetic field B. The field exerts equal and opposite forces on the two arms PQ and RS of the loop. These forces act on the same line, they do not form torque.
Field B is perpendicular to the arm PQ of the loop and exerts a force F1, directed outward to the plane of the loop.
Its magnitude is, F1 = i.a.B
Similarly, the field B exerts an equal force F2 on the arm SR directed into the plane of the paper.
F2 = i.a.B = F1
Thus, the net force on the loop is zero. But there is a torque on the loop due to the pair of forces F1 and F2 which tends to rotate it.
This torqueT= F x perpendicular separation
T = i.a.B.bcosα
α = angle between the magnetic field and the plane of the loop
T= i.B.Acosα as a.b =A area of loop
For a loop having N turns,
T= BiNAcosα
The direction of the torque is along the axis of the loop.
Also T= BiNAsinθ
θ = angle between the magnetic field and the normal to the plane of the loop.
Numericals
4.9 Qn. A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30º with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?
4.13 (a) A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60º with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.
(b) Would your answer change, if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered.)
Qn. Define magnetic dipole moment and write its SI unit. Is it a scalar quantity?
Magnetic dipole moment of a current carrying circular loop m = iNA
Where I = current, N = total turns in loop, A= area of loop
It's SI unit is ampere.aquare metre.
It is a vector quantities shown in the figure.
Qn. Find expression for force between two parallel infinite long straight current carrying conductors. Hence define 1 ampere.
Consider two long conductors ‘a’ and ‘b’ placed parallel, separated by a distance 'd' and carrying currents ia and ib respectively.
The conductor ‘a’ produces, magnetic field Ba at all points along the conductor ‘b’.
Ba=µ02ia/4πd (bye right hand rule, downwards, when the conductors are placed horizontally.)
So a magnetic force acts on conductor b given by
Fba = ibLxBa = ibL.Ba.sin900
Fba = ibL.µ02ia/4πd towards left-side using FLH Rule
Force on unit length of the conductor 'b'
Fba/L=µ02iaib/4πd
The conductor ‘b’ produces, the same magnetic field Bb at all points along the conductor ‘a’ in upward direction.
Bb=µ02ib/4πd
So a magnetic force on length L of the conductor 'a'
Fab = iaLxBb = iaL.Bb.sin900
Fab = iaL.µ02ib/4πd towards right side
Force on unit length of the conductor 'a'
Fab/L=µ02iaib/4πd
Thus Fba = –Fab
We see that currents flowing in the same direction attract each other.
Parallel currents attract, and antiparallel currents repel.
We define 1A as
"The ampere is the value of that steady current which, when maintained in each of the two very long, straight, parallel conductors of negligible cross-section, and placed one metre apart in vacuum, would produce on each of these conductors a force equal to 2 × 10–7 newtons per metre of length."
Numericals
Qn The wires which connect the battery of an automobile to its starting motor carry a current of 300 A (for a short time). What is the force per unit length between the wires if they are 70 cm long and 1.5 cm
apart? Is the force attractive or repulsive?
Qn. Write the value of magnetic field at the midpoint of line joining to very long straight current carrying conductors carrying currents in same direction.
Qn 4.7 Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.
Qn. Find expression for force on a current-carrying conductor kept in a magnetic field.
Consider a rod of uniform cross-sectional area A and length L.
Let the number density of the mobile charge carriers (electrons) = n.
Then the total number of mobile charge carriers in it N = n.volume = n.A.L.
For a steady current i, drift velocity is v
In the presence of an external magnetic field B, Lorentz force on each electron f = q.v × B
the force on conductor is equal to the sum of forces on all moving charges given by F = N.f = (nAL)q.v×B
exchanging direction of velocity with L, we get
F = nqvd.AL×B
but drift current i = n.q.v.A
hence F= i.L×B
here L is a vector of magnitude L, the length of the rod, and with a direction identical to the current i.
Direction of this force can be obtained by Fleming's Left Hand Rule.
Example 4.1 A straight wire of mass 200 g and length 1.5 m carries a current of 2 A. It is suspended in mid-air by a uniform horizontal magnetic field B. What is the magnitude of the magnetic field?
4.5 What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of 30º with the direction of a uniform magnetic field of 0.15 T?
High Order Thinking Question
4.21 A straight horizontal conducting rod of length 0.45 m and mass 60 g is suspended by two vertical wires at its ends. A current of 5.0 A is set up in the rod through the wires.
(a) What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero?
(b) What will be the total tension in the wires if the direction of current is reversed keeping the magnetic field same as before?
(Ignore the mass of the wires.) g = 9.8 m s–2.
Qn Write expression for the Force on a charge moving in electric and magnetic fields: Lorentz Force:
F = q.E + q.v × B
Details:
Suppose a point charge q is moving with a velocity v, in a magnetic field B. Then a force acts on an it. It is given by F = q.v × B If an electric field is also present there, then the force on the charge q is F = q.E + q.v×B ≡ Felectric +Fmagnetic It is called the Lorentz force. Its features: (i) It depends on q, v and B, Force on a negative charge is opposite to that on a positive charge. (ii) The magnetic force q.v × B become zero if velocity and magnetic field are parallel or anti-parallel. The force acts in a (sideways) direction perpendicular to both the velocity and the magnetic field. Its direction is given by the screw rule or right hand rule for vector (or cross) product. (iii) The magnetic force is zero if charge is not moving. Only a moving charge feels the magnetic force.
Qn. Write the SI unit of magnetic field B.
tesla (T)
Details: we have [B] = [F/qv] and the unit of B is N.s/(C.m). This unit is called tesla (T).
Qn what will be the path of a charge particle, projected perpendicular to a uniform magnetic field.
The particle will describe a circle if v and B are perpendicular to each other.
Explanation:
The perpendicular force, q v × B, acts as a centripetal force.
m.v 2/r = q.v.B, which gives
r = m.v / qB
radius of the circle is proportional to velocity
v = ω.r. So,
ω = 2π ν = q B/ m
The time taken for one revolution is
T= 2π/ω ≡ 1/ν = 2πm/qB
Angular frequency does not depend on the speed and radius of the circle
In this case no work is done by the magnetic force on the charge because the force and the velocity are perpendicular. By the work energy theorem it comes out that the change in kinetic energy of the particle will be zero. But it's momentum will change as the direction changes.
4.11 Qn. In a chamber, a uniform magnetic field of 6.5 G (1 G = 10–4 T) is maintained. An electron is shot into the field with a speed of 4.8 × 106 m /s normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit.
(e = 1.6 × 10–19 C, me = 9.1×10–31 kg)
Qn. What will be the path of a charged particle, projected at some angle to a uniform magnetic field.
In this case its path will be helical.
Explanation:
It's velocity will have two components: One parallel to the magnetic field and another perpendicular to the magnetic field.
Due to the component parallel to the magnetic field, the particle shifts along the field direction with uniform speed.
Due to the perpendicular component, it moves and a circle.
Due to the combined effect of these two motions, it's path becomes helical.
The distance moved along the magnetic field in one rotation is called pitch p.
p = v||T = 2πm v|| / q B
Qn. Discuss the magnetic field produced by a straight long solenoid. Use Ampere's circuital law to find the magnetic field inside it.
A solenoid consists turns of insulated copper wire wound closely in the form of a helix.
Circular current in each turn, produces magnetic field along its axis. So all the turns of solenoid produce equal magnetic fields along the axis of the solenoid. This magnetic field B is constant inside the solenoid along the axis; the outside field is negligible because the solenoid is very long.
To find the magnetic field we imagine an rectangular Amperian loop as shown in the figure. The field B is constant on length L inside the solenoid. Other two arms are perpendicular to the field B so B.dl =0. Now using ampere's law
Numericals
4.8 A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of B inside the solenoid near its centre.
Qn. State Ampere's circuital law.
Ampere's circuital law states that
the line integral of the magnetic field around any closed path is equal to permeability constant uo times the total current Ie threading or passing through the closed path . ∮B.dl =µ0.ie
Or
if field B is directed along the tangent to every point on the perimeter L of a closed curve and its magnitude B is constant along the curve , then
B.L = µ0.ie
where I is the net current enclosed by the closed circuit.
Qn. Using ampere's circuital law, find the magnetic field due to an infinite long straight current carrying wire.
The figure shows the circular loop of radius r around an infinite long straight wire carrying current I. As the field lines are circular, the field B at any point of the circular loop is directed along the tangent to the circle at that point. By symmetry the magnitude of field B is same at every point on circumference. So by using ampere 's circuital law
Numericals
4.2 A long straight wire carries a current of 35 A. What is the magnitude of field B at a point 20 cm from the wire?
4.3 A long straight wire in the horizontal plane carries a current of 50 A in north to south direction. Give the magnitude and direction of B at a point 2.5 m east of the wire.
4.4 A horizontal overhead 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?
Qn. Using Biot Savart's law, find expression for magnetic field on the axis of a circular current loop.
Consider a circular loop carrying a steady current i. The loop is placed in the y-z plane with its centre at the origin O and has radius R = a. The x-axis is the axis of the loop. Let x be the distance of a P from the centre O of the loop. The magnitude dB of the magnetic field due to a current element idl is given by the Biot-Savart law
dB =µ0.idlSinθ/4πr2 ........ (1)
Here, r2 = x2 + a2 . Further, any element of the loop will be perpendicular to the displacement vector from the element to the axial point. . Thus, Sinθ = 1
dB =µ0.idl/4πr2 ......... (2)
The direction of dB is shown in fig. It is perpendicular to the plane formed by dl and r. It has an x-component dBx and a component perpendicular to x-axis, dB⊥ . When the components perpendicular to the x-axis are summed over, dB⊥ component due to dl is cancelled by the contribution due to the diametrically opposite dl element, shown in fig. Thus, only the x-component survives. The net contribution along x-direction can be obtained by integrating dBx = dB cos θ over the loop.
cos θ = R/r = a/r .......... (3)
From eq. (2) and (3),
dBx=µ0.idl.cosθ/4πr2 , or
dBx=µ0.idl.a/4πr3 The summation of elements dl over the loop yields 2πa, the circumference of the loop. Thus, the magnetic field at P due to entire circular loop is
B=µ0.2πNia2/4π(a2+x2)3/2......(4)
- Special case - we may obtain the field at the centre of the loop. When x = 0, the above eq. becomes B0 = µ0.2πNi/4πa ........ (5)
- When x2 >>a2 B=µ0.2πNia2/4πx3 But πNia2 =NiA =m B=µ0.2m/4πx3
- It is similar to the electric field at a point on the axis of electric dipole.
Qn. Draw magnetic field lines due to a circular current loop.
The magnetic field lines due to a circular wire form closed loops.
Qn. How can we find direction of magnetic field lines due to a circular current loop?
The direction of the magnetic field is given by (another) right-hand thumb rule given below
Curl the palm of right hand around the circular wire with the fingers pointing in the direction of the current. The right-hand thumb gives the direction of the magnetic field.
Numericals
Two concentric circular coils X and Y of radii 16 cm and 10 cm, respectively, lie in the same vertical plane containing the north to south direction. Coil X has 20 turns and carries a current of 16 A; coil Y has 25 turns and carries a current of 18 A. The sense of the current in X is anticlockwise, and clockwise in Y, for an observer looking at the coils facing west. Give the magnitude and direction of the net magnetic field due to the coils at their centre.
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