# H C VERMA Solutions for Class 12-science Physics Chapter 13 - Magnetic Field due to Current

## Chapter 13 - Magnetic Field due to Current Exercise 249

Using the formulae and , show that the SI units of the magnetic field B and the permeability constant may be written as N/A-m and respectively.

Unit of B

Now, for

Unit of

A current of 10A is established in a long wire along the positive Z-axis. Find the magnetic field at the point (1m,0,0).

Magnetic field due to wire

T

Field is along positive y-axis by the right hand thumb rule.

## Chapter 13 - Magnetic Field due to Current Exercise 250

A copper wire of diameter 1.6mm carries a current of 20A. Find the maximum magnitude of the magnetic field due to this current.

Maximum magnitude of magnetic field will be at the surface of wire i.e., at distance r=0.8mm from center of wire.

mT

A transmission wire carries a current of 100A. What would be the magnetic field B at a point on the road if the wire is 8m above the road?

Magnetic field due to wire

mT

A long, straight wire carrying a current of 1.0A is placed horizontally in a uniform magnetic field T pointing vertically upward. Find the magnitude of the resultant magnetic field at the points P and Q, both situated at a distance of 2.0cm from the wire in the same horizontal plane.

Magnetic field due to the wire is given by

T

Net magnetic field at point P

T

Net magnetic field at point Q

A long, straight wire of radius r carries a current i and placed horizontally in a uniform magnetic field B pointing vertically upward. The current is uniformly distributed over its cross-section. (a) At what points will the resultant magnetic field have maximum magnitude? (b) What will be the minimum magnitude of the resultant magnetic field?

(a)

Magnetic field due to wire is maximum at the surface of wire.

(b) magnetic field due to wire is maximum at the surface

If

So, If then minimum magnetic field

If

then minimum magnetic field is

A long, straight wire carrying a current of 30A is placed in an external, uniform magnetic field of T parallel to the current. Find the magnitude of the resultant magnetic field at a point 2.0cm away from the wire.

Magnetic field due to wire

T

Now, field due to wire and external field are perpendicular to each other.

T

A long, vertical wire carrying a current of 10A in the upward direction is placed in a region where a horizontal magnetic field of magnitude T exists from south to north. Find the point where the resultant magnetic field is zero.

Let at distance r from wire in the west direction net magnetic field is zero.

So,

mm

Figure shows two parallel wires separated by a distance of 4.0cm and carrying equal currents of 10A along opposite directions. Find the magnitude of the magnetic field B at points and.

Magnetic field due to long wire

At point

T

T

T

At point

T

T

T

At point

T

T

T

At point

T

Both fields are perpendicular to each other

T

Two parallel wires carry equal currents of 10A along the same direction and are separated by a distance of 2.0cm. Find the magnetic field at a point which is 2.0cm away from each of these wires.

Magnetic field due to wire P and Q

T

T

Two long, straight wires, each carrying a current of 5A, are placed along the X- and Y-axes respectively. The currents point along the positive directions of the axes. Find the magnetic fields at the points (a) (1m, 1m), (b) (-1m, 1m), (c) (-1m, -1m), (d) (1m, -1m).

(a)

Magnetic field due to wires are of equal magnitude but in opposite direction.

So, resultant field=0.

(b)

Magnetic field due to both wires is in upward direction and is of equal magnitude.

T

(c)

Magnetic field due to wires are of equal magnitude but in opposite direction.

So, resultant field=0.

(d)

Magnetic field due to both wires is in upward direction and is of equal magnitude.

T

Four long, straight wires, each carrying a current of 5.0A, are placed in a plane as shown in figure. The points of intersection form a square of side 5.0cm. (a) Find the magnetic field at the center P of the square. (b) and are points situated on the diagonals of the square and at a distance from P that is equal to the length of the diagonal of the square. Find the magnetic fields at these points.

Magnetic field due to -long wire=

Let magnetic field due to wire (1), (2), (3) and (4) be respectively.

The magnetic field due all four wires at point P is equal in magnitude.

The direction of magnetic fields are.

So, net magnetic field=0

At point

At point

But opposite in direction, so will cancel each other.

But opposite in direction, so will cancel each other.

So, net magnetic field=0

At point

Figure shows a long wire bent at the middle to form a right angle. Show that the magnitudes of the magnetic field at the points P, Q, R and S are equal and find this magnitude.

Magnetic field at the axial point of wire is zero.

Magnetic field due to semi-infinite wire is=

Magnetic field due to wire (1) and wire (2) be and respectively.

At point P

(upward)

(upward)

At point Q

(downward)

(downward)

At point R

(downward)

(downward)

At point S

(upward)

(upward)

Consider a straight piece of length x of a wire carrying a current i. Let P be a point on the perpendicular bisector of the piece, situated at a distance d from its middle point. Show that for d >>x, the magnetic field at P varies as whereas for d<<x, it varies as 1/d.

Here,

Magnetic field at point P

When d>>>x (neglecting x)

When d<<<x (neglecting d)

Consider a long piece of wire which carries a current of 10A. Find the magnitude of magnetic field due to the piece of point which makes an equilateral triangle with the ends of the piece.

Magnetic field at P

T

A long, straight wire carries a current i. Let be the magnetic field at a point P at a distance d from the wire. Consider a section of length l of this wire such that the point P lies on a perpendicular bisector of the section. Let be the magnetic field at this point due to this section only. Find the value of d/l so that differs from by 1%.

Magnetic field due to infinite wire

Magnetic field due to section

Here,

On solving,

Figure shows a square loop ABCD with edge length a. The resistance of the wire ABC is r and that of ADC is 2r. Find the magnetic field B at the center of the loop assuming uniform wires.

Wire ABC and wire ADC are in parallel combination.

So, current will be distributed in inverse ratio of resistance.

Magnetic field due to AB and BC at center of loop

Net field at center=

Figure shows a square loop of edge a made of a uniform wire. A current i enters the loop at the point A and leaves it at the point C. Find the magnetic field at the point P which is on the perpendicular bisector of AB at a distance a/4 from it.

The current will be equally distributed in wire ABC and ADC as both are in parallel combination and have same resistance.

Now,

But in opposite direction hence will cancel each other.

(upward direction)

(downward direction)

Net magnetic field

## Chapter 13 - Magnetic Field due to Current Exercise 251

Consider the situation described in the previous problem. Suppose the current i enters the loop at the point A and leaves it at the point B. Find the magnetic field at the center of the loop.

The current in wire ADCB and wire AB will be in inverse ratio of resistance.

Magnetic field at center of loop

The wire ABC shown in figure forms an equilateral triangle. Find the magnetic field B at the center O of the triangle assuming the wire to be uniform.

The current is equally divided in both sides.

But in opposite direction.

But in opposite direction.

So, net magnetic field=0

A wire of length l is bent in the form of an equilateral triangle and carries an electric current i. (a) Find the magnetic field B at the center. (b) If the wire is bent in the form of a square, what would be the value of B at the center?

(a)

Perpendicular distance of point O from wire PR=

Net magnetic field,

(b)

A long wire carrying a current i is bent to form a plane angle. Find the magnetic field B at a point on the bisector of this angle situated at a distance x from the vertex.

Magnetic field at P due to single wire

Net magnetic field

Find the magnetic field B at the center of a rectangular loop of length l and width b, carrying a current i.

A regular polygon of n sides is formed by bending a wire of total length which carries a current i. (a) Find the magnetic field B at the center of the polygon. (b) By letting, deduce the expression for the magnetic field at the center of a circular current.

Length of each side,

Angle by each side at center

Magnetic field by a side at center

Magnetic field by all sides

Each of the batteries shown in figure has an emf equal to 5V. Show that the magnetic field B at the point P is zero for any set values of the resistances.

Net current in the circuit=0

So, magnetic field is zero at point P.

A straight, long wire carries a current of 20A. Another wire carrying equal current is placed parallel to it. If the force acting on a length of 10cm of the second wire is N, what is the separation between them?

Force per unit length between parallel wires

cm

Three coplanar parallel wires, each carrying a current of 10A along the same direction, are placed with a separation 5.0cm between the consecutive ones. Find the magnitude of the magnetic force per unit length acting on the wires.

On wire (1)

N towards middle wire

On wire (2)

But opposite in direction.

On wire (3)

N towards middle wire

Two parallel wires separated by a distance of 10cm carry currents of 10A and 40A along the same direction. Where a third current should be placed so that it experiences no magnetic force?

Let third wire is placed in middle of both wires at a distance of x from wire carrying current of 10A

Since, no magnetic force

x=2cm

Figure shows a part of an electric circuit. The wires AB, CD and EF are long and have identical resistances. The separation between the neighboring wires is 1.0cm. The wires AE and BF have negligible resistance and the ammeter reads 30A. Calculate the magnetic force per unit length of AB and CD.

Current in each wire will be same as all have same resistance.

So, current= 10A in each wire.

For wire AB

Force due to CD wire + Force due to EF wire.

For wire CD,

Force due to AB is equal to force due to FF wire but in opposite direction.

A long, straight wire is fixed horizontally and carries a current of 50.0A. A second wire having linear mass density is placed parallel to and directly above this wire at a separation of 5.0mm. What current should this second wire carry such that the magnetic repulsion can balance its weight?

Magnetic force between wires=weight of second wire

A

A square loop PQRS carrying a current of 6.0A is placed near a long wire carrying 10A as shown in figure. (a) Show that the magnetic force acting on the part PQ is equal and opposite to that on the part Rs. (b) Find the magnetic force on the square loop.

(a)At distance x from wire consider two small elements of length dx in wire PQ and SR.

Magnetic field is same at both location.

Force on element in PQ is in upward direction and force on element in SR is in downward direction but in opposite direction.

So,

(b) magnetic field at SP by wireT

Force of SP wire=

N (in right direction)

magnetic field at RQ by wireT

Force of SP wire=

N (in left direction)

So, net force

N

A circular loop of one turn carries a current of 5.00A. If the magnetic field B at the center is 0.200mT, find the radius of the loop.

Magnetic field for a circular loop

cm

A current-carrying circular coil of 100 turns and radius 5.0cm produces a magnetic field of T at its center. Find the value of the current.

Magnetic field for a circular loop

mA

An electron makes revolutions per second in a circle of radius 0.5 angstrom. Find the magnetic field B at the center of the circle.

Magnetic field at center

T

A conducting circular loop of radius a is connected to two long, straight wires. The straight wires carry current i as shown in figure. Find the magnetic field B at the center of the loop.

Current in each semi-circle=

Magneti field due to each semicircle is same but in opposite direction. So, net magnetic field is zero.

## Chapter 13 - Magnetic Field due to Current Exercise 252

Two circular coils of radii 5.0cm and 10cm carry equal currents of 2.0A. The coils have 50 and 100 turns respectively and are placed in such a way that their planes as well as the centers coincide. Find the magnitude of the magnetic field B at the common center of the coils if the currents in the coils are (a) in the same sense. (b) in the opposite sense.

Magnetic field due to smaller coil

T

Magnetic field due to bigger coil

T

(a) for same direction

T

(b) for opposite direction

If the outer coil of the previous problem is rotated through 90 about a diameter, what would be the magnitude of the magnetic field B at the center?

Now, magnetic field due to both coils will be in perpendicular direction.

mT

A circular loop of radius 20cm carries a current of 10A. An electron crosses the plane of the loop with a speed of m/s. The direction of motion makes an angle of 30 with the axis of the circle and passes through its center. Find the magnitude of the magnetic force on the electron at the instant it crosses the plane.

r=20cm=0.2m

N

A circular loop of radius R carries a current I. Another circular loop of radius r(<<R) carries a current i and is placed at the center of the larger loop. The planes of the two circles are at right angle to each other. Find the torque acting on the smaller loop.

Magnetic field due to larger loop at center=

Torque acting on smaller loop

A circular loop of radius r carrying a current i is held at the center of another circular loop of radius R (>>r) carrying a current I. The plane of the smaller loop makes an angle of 30 with that of the larger loop. If the smaller loop is held fixed in this position by applying a single force at a point on its periphery, what would be the minimum magnitude of this force?

Magnetic field due to larger loop at center=

Torque acting on smaller loop

Since, smaller loop is at rest

Find the magnetic field B due to a semicircular wire of radius 10.0cm carrying a current of 5.0A as its center of curvature.

Magnetic field due to semi-circle

T

A piece of wire carrying a current of 6.00A is bent in the form of a circular arc of radius 10.0cm, and it subtends an angle of 120 at the center. Find the magnetic field B due to this piece of wire at the center.

T

A circular loop of radius r carries a current i. How should a long, straight wire carrying a current 4i be placed in the plane of the circle so that the magnetic field at the center becomes zero?

Let straight wire is placed at distance x from center of loop.

Since, magnetic field at center is zero.

A circular coil of 200 turns has a radius of 10cm and carries a current of 2.0A. (a) Find the magnitude of the magnetic field at the center of the coil. (b) At what distance from the center along the axis of the coil will the field B drop to half its value at the center? ()

N=200 turns; R=0.1m;I=2A

(a)

mT

(b)

Here,

x=7.66cm

A circular loop of radius 4.0cm is placed in a horizontal plane and carries an electric current of 5.0A in the clockwise direction as seen from above. Find the magnetic field (a) at a point 3.0cm above the center of the loop. (b) At a point 3.0cm below the center of the loop.

(a) magnetic field along axis of a circular loop is given by

T (downward direction)

(b) magnetic field along axis of a circular loop is given by

T (downward direction)

A charge of C is distributed uniformly over a circular ring of radius 20.0cm. The ring rotates about its axis with an angular velocity of 60.0 rad/s. Find the ratio of the electric field to the magnetic field at a point on the axis at a distance of 5.00cm from the center.

C

r=20cm=0.2m

sec

A

Electric field,

Magnetic field,

m/s

A thin but long, hollow, cylindrical tube of radius r carries a current i along its length. Find the magnitude of the magnetic field at a distance r/2 from the surface (a) inside the tube. (b) Outside the tube.

(a) The current enclosed by ampere loop of radius r/2 is zero.

So, magnetic field is also zero.

(b) Magnetic field outside hollow cylindrical tube

A long, cylindrical tube of inner and outer radii a and b carries a current i distributed uniformly over its cross-section. Find the magnitude of the magnetic field at a point (a) just inside the tube. (b) Just outside the tube.

(a) The current enclosed by ampere loop inside the tube is zero. So, magnetic field is also zero.

(b) Consider an ampere loop just outside the tube of radius b.

A long, cylindrical wire of radius b carries a current i distributed over its cross-section. Find the magnitude of the magnetic field at a point inside the wire at a distance a from the axis.

Consider an ampere loop of radius 'a' with center on the axis of cylinder.

Current enclosed=

By ampere's circuital law

A solid wire of radius 10cm carries a current of 5.0A distributed uniformly over its cross-section. Find the magnetic field B at a point at a distance (a) 2cm (b) 10cm and (c) 20cm away from the axis. Sketch a graph of B versus x for 0 < × < 20cm.

(a) Magnetic field inside solid wire of radius 'b' at a distance of 'a' from center of the axis of cylinder.

T

(b) magnetic field at the surface of wire

T

(c) magnetic field at the outside of wire

T

Sometimes we show an idealized magnetic field which is uniform in a given region and falls to zero abruptly. One such field is represented in figure. Using Ampere's law over the path PQRS, show that such a field is not possible.

Consider a point A inside the loop PQRS where B=0.

According to Ampere's circuital law,

If there is current enclosed by the loop PQRS, then magnetic field B cannot be zero.

Whereas, we have taken magnetic field at point A to be zero.

Thus, such field is not possible.

Two large metal sheets carry surface currents as shown in figure. The current through a strip of width dl is Kdl where K is constant. Find the magnetic field at the points P, Q and R.

Magnetic field due to a sheet at a distance 'd' from it having current through a strip of width dl is 'Kdl' can be calculated by ampere's circuital law.

(independent of distance)

At point P and R

Field due to both sheets are equal in magnitude but opposite in direction. Hence, net field is zero.

At point Q

Field due to both sheets are equal and is in same direction.

Consider the situation of the previous problem. A particle having charge q and mass m is projected from the point Q in a direction going into the plane of the diagram. It is found to be describe a circle of radius r between the two plates. Find the speed of the charged particle.

We know that,

The magnetic field B inside a long solenoid, carrying a current of 5.00A, is T. Find the number of turns per unit length of the solenoid.

Magnetic field for solenoid

turns/m

A long solenoid is fabricated by closely winding a wire of radius 0.5mm over a cylindrical nonmagnetic frame so that the successive turns nearly touch each other. What would be the magnetic field B at the center of the solenoid if it carries a current of 5A?

Width of each turn=1mm=m

No. of turns=

T

A copper wire having resistance 0.01ohm in each meter is used to wind a 400-turn solenoid of radius 1.0cm and length 20cm. Find the emf of a battery which when connected across the solenoid will cause a magnetic field of T near the center of the solenoid.

T

No. of turns per unit length

Resistance of the wire

Emf

V=IR

volt

## Chapter 13 - Magnetic Field due to Current Exercise 253

A tightly-wound solenoid of radius a and length l has n turns per unit length. It carries an electric current i. Consider a length dx of the solenoid at a distance x from one end. This contains ndx turns and may be approximated as a circular current i n dx. (a) Write the magnetic field at the center of the solenoid due to this circular current. Integrate this expression under proper limits to find the magnetic field at the center of the solenoid. (b) Verify that if l>>a, the field tends to and if a>>l, the field tends to. Interpret these results.

(a) Magnetic field at the axis due to the circular loop

Here, radius=a distance of the center of the solenoid from the center of circular current=

For the whole solenoid

(b) when l>>>a

So,

If a>>>l

So,

A tightly-wound, long solenoid carries a current of 2.00A. An electron is found to execute a uniform circular motion inside the solenoid with a frequency of rev/s. Find the number of turns per meter in the solenoid.

Frequency of particle in uniform magnetic field

A tightly-wound, long solenoid has n turns per unit length, a radius r and carries a current i. A particle having charge q and mass m is projected from a point on the axis in a direction perpendicular to the axis. What can be the maximum speed for which the particle does not strike the solenoid?

The particle will undergo uniform circular motion. For particle not to strike with solenoid, radius is r/2.

Radius=

A tightly-wound, long solenoid is kept with its axis parallel to a large metal sheet carrying a surface current. The surface current through a width dl of the sheet is Kdl and the number of turns per unit length of the solenoid is n. The magnetic field near the center of the solenoid is found to be zero. (a) Find the current in the solenoid. (b) If the solenoid is rotated to make its axis perpendicular to the metal sheet, what would be the magnitude of the magnetic field near its center?

Magnetic field due to the metal sheet

Magnetic field due to solenoid

(a) Magnetic field is zero.

So,

(b) Now, field due to sheet and solenoid is perpendicular

A capacitor of capacitance 100 is connected to a battery of 20volts for a long time and then disconnected from it. It is now connected across a long solenoid having 4000 turns per meter. It is found that the potential difference across the capacitor drops to 90% of its maximum value in 2.0seconds. Estimate the average the magnetic field produced at the center of the solenoid during this period.

C=100; V=20volt

Q=CV

C

Now, potential drops to 90%

volt

C

Average current flown=

A

Magnetic field at center of solenoid

T

### Kindly Sign up for a personalised experience

- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions

#### Sign Up

#### Verify mobile number

Enter the OTP sent to your number

Change