# H C VERMA Solutions for Class 12-science Physics Chapter 12 - Magnetic Field

## Chapter 12 - Magnetic Field Exercise 230

An alpha particle is projected vertically upward with a speed of km/s in a region where a magnetic field of magnitude 1.0T exists in the direction south to north. Find the magnetic force that acts on the particle.

N towards west.

An electron is projected horizontally with a kinetic energy of 10 keV. A magnetic field of strength T exists in the vertically upward direction. (a) Will the electron deflect towards right or towards left of its motion? (b) Calculate the sideways deflection of the electron in travelling through 1m. Make appropriate approximations.

(a) By Right hand rule the electron will be deflected towards left.

(b) Kinetic energy,

(horizontally)

Magnetic force,

(along west direction)

Time taken to travel 1m in horizontal direction

In this time, deflection of electron in west direction

deflection

deflection=

deflection=

deflection=

deflectionm

deflectioncm

deflectioncm

A magnetic field of T exerts a force of N on a particle having a charge of C and going in the X-Y plane. Find the velocity of the particle.

Let velocity of the particle in X-Y plane

Force on particle,

Comparing,

Velocity,

An experiment's diary reads as follows:" a charged particle is projected in a magnetic field of T. The acceleration of the particle is found to be . The number to the left in the last expression was not readable. What can this number be?

Since, force exerted by magnetic field is always perpendicular to each other.

So,

So, acceleration of particle is .

A 10g bullet having a charge of is fired at a speed of 270m/s in a horizontal direction. A vertical magnetic field of exists in the space. Find the deflection of the bullet due to the magnetic field as it travels through 100m. Make appropriate approximations.

Time to travel 100m distance horizontally,

sec

Force on the bullet

Deflection due to the magnetic field

deflection

deflection=

deflectionm

When
a proton is released from rest in a room, its starts with an initial
acceleration a_{0} towards west. When it is projected towards north
with a speed *v*_{0}, it moves with an
initial acceleration 3a_{0} towards west. Find the electric field and
the maximum possible magnetic field in the room.

Magnetic field does not exert any force on a particle at rest.

So, initially force have been acted by electric field.

(along west)

When projected towards north, force will be exerted by both fields and in same direction as acceleration becomes thrice in same direction (along west).

(into the paper perpendicularly)

Consider a 10cm long portion of a straight wire carrying a current of 10A placed in a magnetic field of 0.1T making an angle of 53° with the wire. What magnetic force does the wire experience?

Force=*i**(l **×** B)*

*F* ≈ 0.08N actin
perpendicular to both wire and magnetic
field.

A current of 2A enters at the corner d of a square frame abcd of side 20cm and leaves at the opposite corner b. A magnetic field B=0.1T exists in the space in a direction perpendicular to the plane of the frame as shown in figure. Find the magnitude and direction of the magnetic forces on the four sides of the frame.

The current will equally be divided at junction d.

So, current along dcb wire and dab wire is 1A.

Since, current, length and magnetic field for all wires is same then magnitude of force is equal.

Magnitude of force,

N

Direction of force can be seen by Right hand rule.

On da and cb wire is towards left.

On dc and ab wire is downward.

## Chapter 12 - Magnetic Field Exercise 231

A magnetic field of strength 1.0T is produced by a strong electromagnet in a cylindrical region of radius 4.0cm as shown in figure. A wire, carrying a current of 2.0A, is placed perpendicular to and intersecting the axis of the cylindrical region. Find the magnitude of the force acting on the wire.

Length of the wire in magnetic field=m

Force of wire,

N

A wire of length l carries a current i along the X-axis. A magnetic field exists which is given as T. Find the magnitude of the magnetic force acting on the wire.

A current of 5.0A exists in the circuit shown in figure. The wire PQ has a length of 50cm and the magnetic field in which it is immersed has a magnitude of 0.20T. Find the magnetic force acting on the wire PQ.

Force,

N upward in the plane of paper.

A circular loop of radius a, carrying a current I, is placed in a two-dimensional magnetic field. The center of the loop coincides with the center of the field. The strength of the magnetic field at the periphery of the loop is B. Find the magnetic force on the wire.

Magnetic field is radially outwards.

Magnetic field and length are perpendicular to each other.

Now,

A hypothetical magnetic field existing in a region is given by , where denotes the unit vector along the radial direction. A circular loop of radius a, carrying a current i, is placed with its plane parallel to the X-Y plane and the center at (0, 0, d). Find the magnitude of the magnetic force acting on the loop.

Magnetic force due to B_{0}cosѲ
field is radially outward which will be counter balanced by each other. So,
force by B_{0}cosѲ field is
zero.

Magnetic force due to B_{0}cosѲ
field is in downward direction.

So, magnetic force F = Bill SinѲ

A rectangular wire-loop of width a is suspended from the insulted pan of a spring balance as shown in figure. A current i exists in the anticlockwise direction in the loop. A magnetic field B exists in the lower region. Find the change in the tension of the spring if the current in the loop is reversed.

Initially, when current is flowing in anticlockwise direction the force on AB and CD wire are counter-balanced while force on BC is in upward direction.

So, initially Tension

Later, when current is flowing in clockwise direction the force will act in downward direction of BC wire.

So, finally Tension

Change in Tension

A current loop of arbitrary shape lies in a uniform magnetic field B. Show that the net magnetic force acting on the loop is zero.

Let the arbitrary loop be rectangle PQRS having length 'l' and breadth 'b'.

Magnetic field B is perpendicular to plane of loop.

Magnitude of force on QP and SR wire is

Both are in opposite direction. So will cancel each other.

Magnitude of force on PS and RQ wire is =
*Bib*

Both are in opposite direction. So will cancel each other.

Therefore, net force acting on the loop is zero.

Prove that the force acting on a current-carrying wire, joining two fixed points a and b in a uniform magnetic field, is independent of the shape of the wire.

Let us consider a semi-circular wire of radius R is placed in uniform magnetic field directed perpendicular of wire.

Considering two element at an angle as shown in figure.

Force acting on each element

Resultant force on both element

Resultant force

(distance between initial and final point)

A semi-circular wire of radius 5.0cm and carries a current of 5.0A. A magnetic field B of magnitude 0.50T exists along the perpendicular to the plane of the wire. Find the magnitude of the magnetic force acting on the wire.

Length of wire

Magnetic field, B=0.5T

Current, i=5A

Force

F=0.25N

A wire, carrying a current i, is kept in the X-Y plane along the curve . A magnetic field B exists in the z-direction. Find the magnitude of the magnetic force on the portion of the wire between x=0 and x=.

Magnetic force depends only upon distance between initial and final point not on shape of wire.

Length of wire

Magnetic force,

A rigid wire consists of a semicircular portion of radius R and two straight sections. The wire is partially immersed in a perpendicular magnetic field B as shown in figure. Find the magnetic force on the wire if it carries a current i.

Magnetic force depends only upon distance between initial and final point of wire.

Length of wire=2R

Force=

F=2iRB

A straight, horizontal wire of mass 10mg and length 1.0m carries a current of 2.0A. What minimum magnetic field B should be applied in the region so that the magnetic force on the wire may balance its weight?

Under equilibrium

Magnetic force=Gravitational force

T

Figure shows a rod PQ of length 20.0cm and mass 200g suspended through a fixed point O by two threads of lengths 20.0cm each. A magnetic field of strength 0.500T exists in the vicinity of the wire PQ as shown in figure. The wires connecting PQ with the battery are loose and exert no force on PQ. (a) Find the tension in the threads when the switch S is open. (b) A current of 2.0A is established when the switch S is closed. Find the tension in the threads now.

(a) Switch 'S' is open

Weight of the rod is balanced by the tension in threads

N

(b) When switch is closed, magnetic force acts in downward direction given by right hand rule.

T=1.25N

## Chapter 12 - Magnetic Field Exercise 232

Two metal strips, each of length l, are clamped parallel to each other on a horizontal floor with a separation b between them. A wire of mass m lies on them perpendicularly as shown in figure. A vertically upward magnetic field of strength B exists in the space. The metal strips are smooth but the coefficient of friction between the wire and the floor is μ. A current i is established when the switch S is closed at the instant t=0. Discuss the motion of the wire after the switch is closed. How far away from the strips will the wire reach?

When switch S is closed, current i flows in the wire. The magnetic force acts in rightward due to which rod moves with constant acceleration on rails. When separated from rails, it is retarded because of frictional force acting in opposite direction of motion.

Let it travels × distance before it stops.

Work done by magnetic force= work done by frictional force

A metal wire PQ of mass 10g lies at rest on two horizontal metal rails separated by 4.90cm. A vertically downward magnetic field of magnitude 0.800T exists in the space. The resistance of the circuit is slowly decreased and it is found that when the resistance goes below 20.0Ω, the wire PQ starts sliding on the rails. Find the coefficient of friction.

When wire is about to slide

Frictional force= Magnetic force

A straight wire of length l can slide on two parallel plastic rails kept in a horizontal plane with a separation d. The coefficient of friction between the wire and the rails is µ. If the wire carries a current i, what minimum magnetic field should exist in the space in order to slide the wire on the rails.

When wire is about to slide

Frictional force=magnetic force

Figure shows a circular wire-loop of radius a, carrying a current i, placed in a perpendicular magnetic field B. (a) Consider a small part dl of the wire. Find the force on this part of the wire exerted by the magnetic field. (b) Find the force of compression in the wire.

(a) Force on small segment of wire,

towards center.

(b) Let a small segment of wire subtends angle at the center of loop.

angle is small]

Suppose that the radius of cross-section of the wire used in the previous problem is r. Find the increase in the radius of the loop if the magnetic field is switched off. The Young's modulus of the material of the wire is Y.

In a magnetic field, the wire is in compressed state and when field is removed it will expand as no compression will be there.

We know that,

The magnetic field existing in a region is given by

A square loop of edge l and carrying a current i, is placed with its edge parallel to the X-Y axes. Find the magnitude of the net magnetic force experienced by the loop.

For wire AB and DC

Magnetic field at distance x is

Length of element=dx

Current in element=i

Since all parameters are same so magnitude of force is also same but opposite in direction. Hence force on wire AB and DC will cancel out each other.

For wire AB

Magnetic field=

Length of wire=l

Current=i

force

(towards left)

For wire BC

Magnetic field

Length of element=l

Current=i

(toward right)

Net force on wire

A conducting wire of length l, lying normal to a magnetic field B, moves a velocity v as shown in figure. (a) Find the average magnetic force on a free electron of the wire. (b) Due to this magnetic force, electrons concentrate at one end resulting in an electric field inside the wire. The redistribution stops when the electric force on the free electrons balances the magnetic force on the free electron balances the magnetic force. Find the electric field developed inside the wire when the redistribution stops. (c) What potential difference is developed between the ends of the wire?

(a) Force on free electron of wire

(b) At equilibrium of electron

Electrostatic force=magnetic force

(c) Potential difference

A current i is passed through a silver strip of width d and width d and area of cross-section A. The number of free electrons per unit volume is n. (a) Find the drift velocity v of the electrons. (b) If a magnetic field B exists in the region as shown in figure, what is the average magnetic force on the free electrons? (c) Due to the magnetic force, the free electrons get accumulated on one side of the conductor along its length. This produces a transverse electric field in the conductor which opposes the magnetic force on the electrons. Find the magnitude of the electric field which will stop further accumulation of electrons. (d) What will be the potential difference developed across the width of the conductor due to the electron-accumulation? The appearance of a transverse emf, when a current-carrying wire is placed in a magnetic field, is called Hall effect.

(a)

(b) Magnetic force,

(c) At equilibrium of electron

Electrostatic force = Magnetic force

(d) potential difference

A particle having a charge C and a mass of g is projected with a speed of m/s in a region having a uniform magnetic field of 0.10T. The velocity is perpendicular to the field. Find the radius of the circle formed by the particle and also the time period.

Radius of circle,

Time period,

sec

A proton describes a circle of radius 1cm in a magnetic field of strength 0.10T. What would be the radius of the circle described by an α - particle moving with the same speed in the same magnetic field?

Radius of the circle

cm

An electron having a kinetic energy of 100eV circulates in a path of radius 10cm in a magnetic field. Find the magnetic field and the number of revolutions per second made by the electron.

Relation between kinetic energy andradius of circle

T

Number of revolutions per second,

Hz

Protons having kinetic energy K emerge from an accelerator as a narrow beam. The beam is bent by a perpendicular magnetic field so that it just misses a plane target kept at a distance l in front of the accelerator. Find the magnetic field.

Proton just misses the target means

Radius<

## Chapter 12 - Magnetic Field Exercise 233

A charged particle is accelerated through a potential difference of 12kV and acquires a speed of m/s. It is then injected perpendicularly into a magnetic field of strength 0.2T. Find the radius of the circle described by it.

Work done by potential difference

Now, radius of the circle

cm

Doubly-ionized helium ions are projected with a speed of 10 km/s in a direction perpendicular to a uniform magnetic field of magnitude 0.1T. Find (a) the force acting on an ion, (b) the radius of the circle in which it circulates and (c) the time taken by an ion to complete the circle.

(a) Force=

N

(b) Radius of the circle,

m

(c) Time period

sec

A proton is projected with a velocity of m/s perpendicular to a uniform magnetic field of 0.6T. find the acceleration of the proton.

Force on the proton

(a) An electron moves along a circle of radius 1m in a perpendicular magnetic field of strength 0.50T. What would be its speed? Is it reasonable? (b) If a proton moves along a circle of same radius in the same magnetic field, what would be its speed?

(a)

m/s

It is not reasonable as its speed is more than speed of light.

(b) For proton

m/s

A particle of mass m and positive charge q, moving with a uniform velocity v, enters a magnetic field B as shown in figure. (a) Find the radius of the circular arc it describes in the magnetic field. (b) Find the angle subtended by the arc at the center. (c) How long does the particle stay inside the magnetic field? (d) Solve the three parts of the above problem if the charge q on the particle is negative.

(a) Magnetic force = centripetal force

(b) Angle subtended at center of circle=

(c) Time period

(d) If charge is negative.

(i) Radius

Magnetic force = centripetal force

(ii) Angle subtended at center

(iii) Time period

A particle of mass m and charge q is projected into a region having a perpendicular magnetic field B. Find the angle of deviation of the particle as it comes out of the magnetic field if the width d of the region is very slightly smaller than

When width of the magnetic field is less than the radius of the circle of particle it will comes and if width of magnetic field is more than radius of circle of particle it will not come out.

(a)

(b)

(c)

It will not come out from forward direction but will describe a semicircle deviation = π

A narrow-beam of singly-charged carbon ions, moving at a constant velocity of m/s, is sent perpendicularly in a rectangular region having uniform magnetic field B=0.5T. It is found that two beams emerge from the field in the backward direction, the separations from the incident beam being 3.0cm and 3.5cm. Identify the isotopes present in the ion beam. Take the mass of an ion=kg, where A is the mass number.

Distance between incident rays and emergent rays=2R

For first beam cm

For second beam cm

Radius of the circle is given by

Mass of 1^{st} particle

kg

amu

So,

The two isotopes are of carbon .

Fe^{+}
ions are accelerated through a potential difference of 500V and are injected
normally into a homogeneous magnetic field B of strength 20.0mT. Find the
radius of the circular paths followed by the isotopes with mass number 57 and
58. Take the mass of an ion=kg where A is the mass number.

Work done= chargepotential difference

For isotopes with mass number 57

m

cm

For isotopes with mass number 58

m

cm

A narrow beam of singly-charged potassium ions of kinetic energy 32keV is injected into a region of width 1.00cm having a magnetic field of strength 0.500T as shown in figure. The ions are collected at a screen 95.5cm away from the field region. If the beam contains isotopes of atomic weights 39 and 41, find the separation between the points where these isotopes strike the screen. Take the mass of a potassium ion=kg where A is the mass number.

Kinetic energy,

For K-39

m/s

Time to cross magnetic field

sec

Now acceleration in the magnetic field region

Displacement along y-direction in field

m

Velocity in the vertical direction;

m/s

Time to reach the screen=sec

During this time, distance travelled in vertical direction

=speed × time

m

Net displacement from horizontal=

m

Similarly, for K-41

Net displacement from horizontal=m

Net gap=mm

Electrons are emitted with negligible speed from an electron gun are accelerated through a potential difference V along the X-axis. These electrons emerge from a narrow hole into a uniform magnetic field B directed along this axis. However, some of the electrons emerging from the hole make slightly divergent angles as shown in figure. Show that these paraxial electrons are focused on the X-axis at a distance

Work done= charge × potential difference

The electrons which are at slightly divergent angles will follow helical path.

Time period to complete one revolution of helical path

In this time period, distance travelled by electron along x-axis will be equal to pitch.

Pitch=

Pitch=

Figure shows a convex lens of focal length 12cm lying in a uniform magnetic field B of magnitude 1.2T parallel to its principal axis. A particle having a charge C and mass kg is projected perpendicular to the plane of the diagram with a speed of 4.8m/s. The particle moves along a circle with its center on the principal axis at a distance of 18cm from the lens. Show that the image of the particle goes along a circle and find the radius of that circle.

Radius of the circle for object

cm

Now, magnification

cm negative sign shows that image is inverted.

## Chapter 12 - Magnetic Field Exercise 234

Two particles, each having mass m are placed at a separation d in a uniform magnetic field B as shown in figure. They have opposite charges of equal magnitude q. At time t=0, the particles are projected towards each other, each with a speed v. Suppose the Coulomb force between the charges is switched off. (a) find the maximum value of the projection speed so that the two particles do not collide. (b) What would be the minimum and maximum separation between the particles if ? (c) At what instant will a collision occur between the particles if ? (d) Suppose and the collision between the particles is completely inelastic. Describe the motion after the collision.

(a) for particles not to collide

d>(R+R)

d>2R

Maximum speed of particles

(b)

Radius of each circle=

Minimum separation between particles=d-2R

=

Maximum separation between particles=d+2R

(c) Radius od the particle will be

The particles will collide after travelling distance in horizontal direction.

Angle rotated by the particle till collision occurs

Time taken=

(d)

By momentum of conservation,

along x-axis,

along y-axis,

So, particle moves with constant speed along the straight line in upward direction after collision as it becomes chargeless.

A uniform magnetic field of magnitude 0.20T exists in space from east to west. With what speed should a particle of mass 0.010g and having a charge C be projected from south to north so that it moves with a uniform velocity?

Particle will move with constant velocity if

Magnetic force=gravitational force

m/s

A particle moves in a circle of diameter 1.0cm under the action of a magnetic field of 0.40T. An electric field of 200 V/m makes the path straight. Find the charge/mass ratio of the particle.

Radius of the circle=cm

If the particle moves with constant velocity

Electrostatic force= magnetic force

C/Kg

A proton goes undeflected in a crossed electric and magnetic field (the fields are perpendicular to each other) at a speed of m/s. The velocity is perpendicular to both the fields. When the electric field is switched off, the proton moves along a circle of radius 4.0cm. Find the magnitudes of the electric and the magnetic fields. Take the mass of proton kg.

Initially,

When electric field is switched off, the proton will move in a circle due to force by magnetic field.

magnetic field, B=0.05T

electric field, N/C

A particle having a charge of and a mass of kg is projected with a speed of 1.0km/s in a magnetic field of magnitude 5.0mT. The angle between the magnetic field and the velocity is . Show that the path of the particle will be a helix. Find the diameter of the helix and its pitch.

C

kg

m/s

T

Component of velocity along magnetic field=

Component of velocity perpendicular to magnetic field=

Radius of the particle,

m

Diameter=2R=2(0.18) =0.36m=36cm

Pitch=

cm

A proton projected in a magnetic field of 0.020T travels along a helical path of radius 5.0cm and pitch 20cm. Find the components of the velocity of the proton along and perpendicular to the magnetic field. Take the mass of the proton=kg.

m=kg

B=0.02T

R=5cm=m

P=20cm=m

Let velocity of the particle perpendicular and parallel to magnetic field be respectively.

Radius of helical path

m/s

Pitch of helical path

m/s

A particle having a mass m and charge q is released from the origin in a region in which electric field and magnetic field are given by and . Find the speed of the particle as a function of its z-coordinate.

Let velocity of the particle after it travelled distance 'z' along z-axis.

Force on the particle

Since particle is in y-z plane only, So

Force along z-axis

Also,

An electron is emitted with negligible speed from the speed from the negative plate of a parallel plate capacitor charged to a potential difference V. The separation between the plates is d and a magnetic field B exists in the space as shown in figure. Show that the electron will fail to strike upper plate if

Electric field setup between the plates of a capacitor,

Work done by electric field = change in kinetic energy

The electron will move in circular path of radius,

Since electron doesn't strike upper plate

A rectangular coil of 100 turns has length 5cm and width 4cm. It is placed with its plane parallel to a uniform magnetic field and a current of 2A is sent through the coil. Find the magnitude of the magnetic field B, if the torque acting on the coil is 0.2N-m.

N=100 turns

I=2A

N-m

We know that,

B=0.5T

A 50-turn circular coil of radius 2.0cm carrying a current of 5.0A is rotated in a magnetic field of strength 0.20T. (a) What is the maximum torque that acts on the coil? (b) In a particular position of the coil, the torque acting on it is half of this maximum. What is the angle between the magnetic field and the plane of the coil?

(a)

N-m

(b) Let angle between magnetic field and plane of coil be

A rectangular loop of sides 20cm and 10cm carries a current of 5.0A. A uniform magnetic field of magnitude 0.20T exists in parallel to the longer side of the loop. (a) What is the force acting on the loop? (b) What is the torque acting on the loop?

(a) Force acting on both longer sides is zero as angle between length of wire and magnetic field is zero.

Force on shorter sides are equal in magnitude but opposite in direction. Hence will cancel out each other.

So, net force on loop is zero.

(b) torque,

N-m

A circular coil of radius 2.0cm has 500 turns in it and carries a current of 1.0A. Its axis makes an angle of 30° with the uniform magnetic field of magnitude 0.40T that exists in the space. Find the torque acting on the coil.

N=500 turns

I=1A

Torque,

N

A circular loop carrying a current i has wire of total length L. A uniform magnetic field B exists parallel to the plane of the loop. (a) Find the torque on the loop. (b) If the same length of the wire is used to form a square loop, what would be the torque? Which is larger?

(a) Length of wire=circumference of circle

Area of circular loop

Torque,

(b) Length of the wire= perimeter of square of side 'a'

L=4a

Area of square=

Torque,

Torque on circular loop is larger than square loop.

A square coil of edge l having n turns carries a current i. It is kept on a smooth horizontal plate. A uniform magnetic field B exists in a directional parallel to an edge. The total mass of the coil is M. What should be the minimum value of B for which the coil will start tipping over?

Torque by the magnetic field

For the coil to start tipping over

torque due to gravitational force

Minimum magnetic field

Consider a nonconducting ring of radius r and mass m which has a total charge q distributed uniformly on it. The ring is rotated about its axis with an angular speedω. (a) Find the equivalent electric current in the ring, (b) find the magnetic moment of the ring. (c) Show that where l is the angular momentum of the ring about its axis of rotation.

(a) Current,

(b) magnetic moment,

(c) angular momentum

Putting this value in magnetic moment expression

## Chapter 12 - Magnetic Field Exercise 235

Consider a nonconducting plate of radius r and mass m which has a charge q distributed uniformly over it. The plate is rotated about its axis with an angular speedω. Show that the magnetic moment µ and the angular momentum l of the plate are related as .

Consider a ring at distance x from center and of width dx

Charge on the elemental ring

Current due to rotation of this elemental ring

Magnetic moment due to elemental ring

Total magnetic moment

Angular momentum

Divide (1) and (2)

Consider a solid sphere of radius r and mass m which has a charge q distributed uniformly over its volume. The sphere is rotated about a diameter with an angular speed ω. Show that the magnetic moment and the angular momentum l of the sphere are related as .

Consider a disc at distance x from center of sphere of thickness 'dx'

Consider an elemental ring at distance y of thickness 'dy' from center of disc

Charge on elemental ring

Current due to elemental ring

Magnetic moment due to elemental ring

Magnetic moment for complete sphere

Angular momentum,

So,

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