## BSEB Bihar Board Class 12 Physics Solutions Chapter 4 Moving Charges and Magnetism

### NCERT Solutions For Class 12 Physics Moving Charges and Magnetism

Question 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?

Answer:

Here, n = number of turns in the coil = 100

r = radius of each turn of the coil = 8.0 cm = 8 x 10^{-2} m.

I = current in the circular coil = 0.40 A.

Let B = magnitude of magnetic field at the centre of coil = ?

Using the relation,

B = \(\frac{\mu_{0}}{4 \pi} \cdot \frac{2 \pi \mathrm{n} \mathrm{l}}{\mathrm{r}}\) we get

B = 10^{-7} x \(\frac{2 \times 3.14 \times 100 \times 0.40}{8 \times 10^{-2}}\)

= 3.14 x 10^{-4}T

Question 2.

A long straight wire carries a current of 35 A. What is the magnitude of the field Bata point 20 cm from the wire?

Answer:

Here, I = current flowing in the straight wire = 35 A.

a = distance of the point from the straight wire = 20 cm

B = magnitude of the magnetic field = ?

Using the relation,

B = \(\frac{\mu_{0}}{4 \pi} \cdot \frac{2 \pi \mathrm{n} \mathrm{l}}{\mathrm{r}}\) weget

B = 10^{-7} x \(\frac{2 \times 35}{20 \times 10^{-2}}\) = 3.5 x 10^{-5} T

Question 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.

Answer:

Here, I = current in the long wire from north to south direction = 50 A.

a = perpendicular distance of the point (P say) from the wire = 2.5 m.

Let B be the magnitude of the field at P.

.’. Using the relation,

B = \(\frac{\mu_{0}}{4 \pi} \cdot \frac{2 \pi \mathrm{n} \mathrm{l}}{\mathrm{r}}\) we get

B = 10^{-7} x \(\frac{2 \times 50}{2.5}\) = 0.4 x 10^{-5} T

= 4 x 10^{-6} T

Direction of \(\overrightarrow{\mathrm{B}}\) :

Applying right hand thumb rule, we find that the direction of \(\overrightarrow{\mathrm{B}}\) at the observation point is towards the east i.e., If we assume that NS direction lies in the plane of paper, then \(\overrightarrow{\mathrm{B}}\) will be perpendicular to the plane of paper and in the vertically upward direction,

Question 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?

Answer:

Here, I = current in the power line = 90 A in east to west direction.

a = 1.5 m.

Let B = magnetic field at point P (say) at a distance 1.5 m below the line.

∴ Using the relation,

B = \(\frac{\mu_{0}}{4 \pi} \cdot \frac{2 \pi \mathrm{n} \mathrm{l}}{\mathrm{r}}\), we get

B = 10^{-7} x \(\frac{2 \times 90}{1.5}\) = 1.2 x 10^{-5} T

Direction of \(\overrightarrow{\mathrm{B}}\) :

According to right hand thumb rule, the direction of the magnetic field is towards the south at the point of observation.

Question 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 30° with the directiorf of a uniform magnetic field at 0.15 T?

Answer:

Here, I = Current in the wire = 8A.

θ = angle made bv the wire with the uniform magnetic field

\(\overrightarrow{\mathrm{B}}\) = 30°

B = 0.15 T

Let F be the magnetic force per unit length on the wire = ?

∴ Using equation

F = BI sin θ , we get

F = 0.15 x 8 x sin 30° (Nm^{-1})

= 0.15 x 8 x \(\frac {1}{2}\)

= 0.15 x 4 = 0.60 Nm^{-1}.

Question 6.

A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire?

Answer:

Here, l = length of the wire = 3.0 cm = 3 x 10^{-2} m.

I = current in the wire = 10 A.

B = magnetic field inside the solenoid = 0.27 T

0 = angle made by the wire with the axis of the solenoid = 90°

Let F = magnitude of the magnetic force on the wire.

∴ Using eqn, F = BIZ sin θ, we get

F = BI l sin 90° = BIl

= 0.27 x 10 x 3 x 10^{-2}

= 8.1 x 10^{-2} N.

Direction of F:

Direction of \(\overrightarrow{\mathrm{F}}\) is found using Fleming’s left hand rule.

Question 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.

Answer:

Here, Let I_{1}, and I_{2} be the currents flowing in the straight long and parallel wires A and B respectively.

∴ I_{1} = 8.0 A, I_{2} = 5.0 A flowing in the same direction.

r = distance between A and B = 4.0 cm = 4 x 10^{-2} m.

If F’ be the force per unit length on wire A, then using

F’ = \(\frac{\mu_{0}}{4 \pi} \cdot \frac{2 \pi \mathrm{n} \mathrm{l}}{\mathrm{r}}\) we get

F’ = 10^{-7} x \(\frac{2 \times 8 \times 5}{4 \times 10^{-2}}\) Nm^{-1}

= 20 x 10^{-5} Nm^{-1}.

If F be the force on a section of length 10 cm of wire A, then

F = F’ x l (Here, l = 10 x 10^{-2} m)

= 20 x 10^{-5} x 10 x 10^{-2} N = 2 x 10^{-5}N.

Direction of \(\overrightarrow{\mathrm{F}}\) : As the current in the two parallel wires is flowing in the same direction, so the force will be attractive normal to A towards B.

Question 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.

Answer:

Here, l = length of the long solenoid = 80 cm = 80 x 10^{-2} m.

n’ = no. of turns in each layer = 400

n = no. of layers in the solenoid =5.

N = total number of turns in the solenoid = n’n = 400 x 5 = 2000.

I = current flowing in the solenoid = 8.0 A.

D = Diameter of the solenoid = 1.8 cm = 1.8 x 10^{-2} m.

Let B be the magnitude of the magnetic field inside the solenoid near its centre = ?

\(\frac{\mu_{0} \mathrm{NI}}{l}\) = \(\frac{4 \pi \times 10^{-7} \times 2000 \times 8}{80 \times 10^{-2}}\) T

= 8π x 10^{-3} T = 2.5 x 10^{-2} T.

Question 9.

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 30° with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?

Answer:

Here, a = side of square coil = 10 cm = 10 x 10^{-2} m.

N = total number of turns in the coil = 20

I = current flowing in the coil = 12 A.

B = magnitude of uniform magnetic field = 0.80 T.

θ = angle made by the normal to the plane of coil with B = 30°

∴ A = area of coil = a^{2} = (10 x 10^{-2}) m^{2}.

τ = magnitude of the torque experienced by the coil = ?

Using the formula,

τ = NIA B sin θ, we get

τ = 20 x 12 x (10 x 10^{-2})^{2} x 0.80 x sin 30°

= 20 x 12 x 10^{-2} x 0.80 x \(\frac {1}{2}\)

= 0.96 Nm.

Question 10.

Two moving coil meters, M_{1} and M_{2} have the following particulars:

R_{1} = 10 Ω, N_{1} = 30,

A_{1} = 3.6 x 10-3 m2, B_{1} = 0.25 T

R_{2} = 14 Ω, N_{2} = 42,

A_{2} = 1.8 x 10^{-3} m^{2}, B_{2} = 0.50 T

(The spring constants are identical for the two meters). Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M_{2} and M_{1}.

Answer:

Here, R_{1} = 10 Ω, N_{1} = 30, A_{1} = 3.6 x 10^{-3} m^{2}, B_{1} = 0.25 T for coil M_{1}

R_{2} = 14 Ω, N_{2} = 42, A_{2} = 1.8 x 10^{3} m^{2}, B_{2} = 0.50 T for coil M_{2}.

We know that current sensitivity and voltage sensitivity are given by the formulae.

Current sensitivity = \(\frac {NBA}{K}\)

and Voltage sensitivity = \(\frac {NBA}{KR}\)

Here k_{1} = k_{2} for the two coils = k (say)

∴ Current sensitivity for M_{1} is given by = N_{1} B_{1}A_{1}/k

and for M_{2} = N_{2} B_{2} \(\frac{\mathrm{A}_{2}}{\mathrm{k}}\)

(a) Current resistivity ratio for M_{2} and M_{1} is given by

Now voltage resistivity for

M_{1} is = \(\frac{N_{1} B_{1} A_{1}}{k R_{1}}\) and for M_{2} = \(\frac{\mathrm{N}_{2} \mathrm{~B}_{2} \mathrm{~A}_{2}}{\mathrm{kR}_{2}}\)

(b) Voltage sensitivity ratio for M_{2} and M_{1} is

Question 11.

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 x 10^{6} ms^{-1} normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit, (e = 1.6 x 10^{-19} C, m_{e} = 9.1 x 10^{-31} kg)

Answer:

Here, B = 6.5 G = 6.5 x 10^{-4} T = uniform magnetic field.

υ = speed of the electron shot normal to \(\overrightarrow{\mathrm{E}}\) = 4.8 x 10^{6} ms^{-1}

e = charge on an electron = 1.6 x 10^{-19} C.

m_{e} = mass of an electron,

=9.1 x 10^{31}kg.

θ = 90° C

If r be the radius of the circular orbit, then the force F on the moving electron due to the magnetic field is given by

F = e υ B sin θ .

The direction of the force is perpendicular to \(\overrightarrow{\mathrm{v}}\) as well as to \(\overrightarrow{\mathrm{B}}\) , so this force will only change the direction of motion of the electron without affecting its speed and this force provides the necessarily centrepetal force to the moving electron, so the electron will move on the circular path. If r be the radius of the circular path followed by the electron, then

Question 12.

In Exercise 4.11 obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.

Answer:

Let v = Frequency of revolution of electron in the orbit = ?

Again using the formula,

From equation (1), we see that the frequency of revolution of the electron does not have the speed of the electron, so v does not depend on the speed of the electron.

Question 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 magnetude 1.0 T. The field lines make an angle of 60° 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.)

Answer:

Here, N = no. of turns in the circular coil = 30

r = radius of the ciruclar coil = 8.0 cm = 8 x 10^{-2} m.

I = current in the circular coil = 6.0 A.

B = uniform horizontal magnetic field = 1.0 T.

θ = angle between \(\overrightarrow{\mathrm{B}}\) and the normal to the plane of coil = 60°

∴ area of circular coil, A = πr^{-2} = 3.14 x (8 x 10^{-2})^{2} m^{2}

= 3.14 x 64 x 10^{-4} m^{2} = 2.01 x 10^{-2} m^{2}.

(a) Magnitude of torque acting on the current carrying coil due to the magnetic field is given by

To prevent the coil from rotating, a torque equal and opposite to x has to be applied

∴ Torque required = τ = 3.133 Nm.

(b) No. Since the torque on a planar loop is independent of its shape in case the area of the loop is kept same; hence the torque will remain unchanged.

Question 14.

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.

Answer:

For coil X:

Here, a = radius of coil = 16 cm = 16 x 10^{-2} m.

n = no. of turns in the coil = 20

I = current in the coil = 16 A (anticlockwise)

B_{x} = The magnetic field at the centre at the coil X = ?

Using formula,

B = \(\frac{\mu_{0}}{4 \pi} \cdot \frac{2 \pi n I}{Q}\) …….(1)

Weget

B = \(\frac{\mu_{0}}{4 \pi} \cdot \frac{2 \pi n I}{9}\)

= \(\frac{10^{-7} \times 2 \times \pi \times 20 \times 16}{16 \times 10^{-2}}\) = 4π x 10^{-4} T

For an observer looking at the coil X facing west, the current is flowing in the anticlockwise direction, so the direction of B is towards east.

For coil Y:

Here, a = 10 cm = 10 x 10^{-2} m

n = 25

I = 18 A (clockwise)

B_{y} = magnetic field at the centre of the coil Y = ?

Using eqn (1), we get

B_{y} = \(\frac{\mu_{0}}{4 \pi} \cdot \frac{2 \pi \mathrm{nI}}{\mathrm{a}}\)

= 10^{-7} x \(\frac{2 \pi \times 25 \times 18}{10 \times 10^{-2}}\) T

= 9π x 10^{-4}T.

For an observer looking at the coil Y facing west, the current is flowing in the clockwise direction, so the direction of field B_{y} is towards west. ‘

Now as B_{x} and B_{y} act along the same line in the opposite direction, so the resultant magnetic field will act in the direction of bigger vector.

So if \(\overrightarrow{\mathrm{E}}\) be the net magnetic field due to the two coils at their centre, then

\(\overrightarrow{\mathrm{E}}\) = \(\overrightarrow{\mathrm{r}_{\mathrm{y}}}\) + \(\overrightarrow{\mathrm{r}_{\mathrm{x}}}\)

= 9π x 10^{-4} (west) + 471 x 10^{-4} (east)

= (9π x 10^{-4} – 4π x 10^{-4}) west

= 5π x 10^{-4} T towards west

= 5 x 3.14 x 10^{-4} T towards west

= 1.6 x 10^{-4}T towards west.

Question 15.

A magnetic field of 100 G (1G = 10^{-4} T) is required which is uniform in a region of linear dimension about 10 cm and area of cross-section about 10^{-3} m^{2}. The maximum current-carrying capacity of a given coil of wire is 15 A and the number of turns per unit length that can be wound round a core is at most 1000 turns m^{-1}. Suggest some appropriate design particulars of a solenoid for the required purpose. Assume the core is not ferromagnetic.

Answer:

Here, B = magnetic field = 100 G = 100 x 10^{-4} = 10^{-2} T,

I_{max} maximum current carried by the coil = 15 A.

n = number of turns per unit length = 1000 turns m^{-1} = 10 turns m^{-1} = 10 turns / cm.

l = length of linear region = 10 cm

A = area of cross-section = 10^{-3} m^{2}.

To produce a magnetic field in the above mentioned region, a solenoid can be made so that well within the solenoid, the magnetic field is uniform. To do so, we may take the length L of the solenoid 5 times the length of the region and area of the solenoid 5 times the area of region. ∴ L = 5l = 50 cm = 0.5

and A = 59 = 5 x 10^{-3} m^{2}

∴ If r be the radius of the solenoid, then

πr^{2} = A = 5 x 10^{-3}

r = \(\sqrt{\frac{5 \times 10^{-3}}{3.14}}\) = 0.04m = 4cm

Also let us wind 500 turns on the coil so that the number of turns per m is

n’ = \({\frac{500}{0.5}\) = 1000 turns m^{-1},

∴ Using formula, μ_{0}nl = B, we get

I = \(\frac{\mathrm{B}}{\mu_{0} \mathrm{n}}\) = \(\frac{10^{-2}}{4 \pi \times 10^{-7} \times 1000}\) = 7.96 A ≈ 8A

So, a current of 8A can be passed through it to produce a uniform magnetic field of loo G in the region. But this is not a unique way. If we wind 300 turns on the solenoid, then number of turns is n = \({\frac{300}{0.5}\) = 600 per m.

∴ I = \(\frac{\mathrm{B}}{\mu_{0} \mathrm{n}}\) = \(\frac{10^{-2}}{4 \pi \times 10^{-7} \times 600}\) = 13.3A

i.e. a current of 10 A can be passed Through it to produce the magnetic field of 100 G

Similary if no .of turns = 400

then

n = \(\frac{400}{0.5}\) = 800 per m.

I = \(\frac{B}{\mu_{0} n}\) = \(\frac{10^{-2}}{4 \pi \times 10^{-7} \times 800}\) = 9.95 A

i.e. a current of 10 A can be passed ≈ 10A

Through it to produce B = 100 G

Thus we may achieve the result in a number of ways.

Question 16.

for a circular coil of radius R and N turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distance x from its centre is given by,

B = \(\frac{\mu_{0} I R^{2} N}{2\left(x^{2}+R^{2}\right)^{3 / 2}}\)

(a) Show that this reduces to the familiar result for the field at the centre of the coil.

(b) Consider two parallel co-axial circular coils of equal radius R, and number of turns N, carrying equal currents in the same direction, and separated by a distance R. Show that the field on the axis around the mid-point between the coils is uniform over a distance that is small as compared to R, and is given by,

B = 0.72 \(\mu_{0} \text { NI }\) , approximately.

[Such an arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils\.

Answer:

Here the magnetic field at a point on the axis of the circular coil ai a distance x is given by

B = \(\frac{\mu_{0} N I R^{2}}{2\left(x^{2}+R^{2}\right)^{\frac{3}{2}}}\) ……(1)

where R = radius of circular coil.

(a) At the centre of the coil, x = 0.

∴ from equation (1), we get

which is the required familiar expression for the magnetic field at the centre of the coil.

(b) Let R be the radius of each coaxial circular coils.

N = number of turns in each coil.

R = distance between the two coils = CD.

Let 2d < Rbe the small distance around the mid-point of CD i.e. O, where we want to show that the field is uniform.

Let I = Current flowing through the two coils in the same direction. We know that the magnetic field at a point on the axis of the coil at a distance x from its centre is given by

Let B_{1} and B_{2} be the magnetic fields at point D’ due to the coils (1) and (2) respectively on their axes.

Putting x = \(\frac {R}{2}\) + d for (1) coil and x = \(\frac {R}{2}\) – d for coil (2) in

equation (i), we get

If B be the total magnetic field at P due to the two coils, then

B = B_{1} + B_{2}

B_{1} and B_{2} are added up because both act in the same direction.

Meglecting d^{2} and d^{2}<<R^{2}

Using Binomial expression theorem, we get

= 0.72. \(\frac{\mu_{0} \mathrm{NI}}{\mathrm{R}}\) which is independent of the small distance and hence is uniform.

Hence proved.

Question 17.

A toroid has a core (non-ferromagnetic) of inner radius 25 cm and outer radius 26 cm, around which 3500 turns of a wire are wound. If the current in the wire is 11 A, what is the magnetic field (a) outside the toroid, (b) inside the core of the toroid, and (c) in the empty space surrounded by the toroid.

Answer:

Here, inner radius r_{1} = 25 cm = 25 x 10^{-2} m

outer radius, r_{2} = 26 cm = 26 x 10^{-2} m.

N = Totahnumber of turns = 3500

I = Current in the wire = 11 A

Let r be the mean radius of the toroid

∴ r = \(\frac{r_{1}+r_{2}}{2}\) = \(\frac{(25+26) \times 10^{-2}}{2}\)

= 25.5 x 10^{-2} m.

Total length of the toroid = circumference of the toroid

= 2πr = 2 x π x 25.5 x 10^{-2} m = 51 π x 10^{-2} m

If n be the number of turns per unit length of toroid, then

n = \(\frac{N}{2πr}\) = \(\frac{3500}{51 \pi \times 10^{-2} \mathrm{~m}^{-1}}\)

(a) Let B_{0} be the magnetic field outside the toroid = ?

The field is non-zero only inside the core surrounded by the windings of the toroid, so the magnetic field outside the toroid is zero i.e B_{0} = 0

(b) Let B_{1} be the magnetic field inside the core of the toroid =?

B_{1} = μ_{0} nI

= 4π x 10^{7} x \(\frac{3500}{51 \pi \times 10^{-2}}\) x 11

= 0.0302 T = 3.02 x 10^{-2} T

(c) The magnetic field in the empty space surrounded by the toroid

is also zero as the field is non zero only inside the core surrounded by the windings of the toroid.

Question 18.

Answer the following questions

(a) A magnetic field that varies in magnitude from point to point but has a constant direction (east to west) is set up in a chamber. A charged particle enters the chamber and travels undeflected along a straight path with constant speed. What can you say about the initial velocity of the particle?

(b) A charged particle enters an environment of a strong and non-uniform magnetic field varying from point to point both in magnitude and direction, and comes out of it following a complicated trajectory. Would its final speed equal the initial speed if it suffered no collisions with the environment?

(c) An electron travelling west to east enters a chamber having a uniform electrostatic field in north to south direction. Specify the direction in which a uniform magnetic field should be set up to prevent the electron from deflecting from its straight line path.

Answer:

(a) We know that the force on a charged particle moving inside the magnetic field is given by :

\(\overrightarrow{\mathrm{F}}\) = q \((\vec{v} \times \overrightarrow{\mathrm{B}})\)

The charged particle will travel undeflected i.e. the force on it will be zero if \(\vec{v} \times \overrightarrow{\mathrm{B}}\) is equal to zero.

\(\vec{v} \times \overrightarrow{\mathrm{B}}\) can be zero either the initial velocity \(\overrightarrow{\mathrm{υ}}\) is parallel or antiparallel to \(\overrightarrow{\mathrm{B}}\)

(b) Yes, its final speed would be equal to the initial speed if it suffered no collision with the environment. This,is because the magnetic field exerts force on the charged particle which is always perpendicular to its motion, so this force can change the direction of motion of the velocity \(\overrightarrow{\mathrm{υ}}\) and not its magnitude.

(c) The electrostatic field is directed towards south. Since the electron is negativity charged particle, therefore the electrostatic field shall exert a force directed towards north. So, if the electron is to be prevented from deflection from straight path, the magnetic force on the electron should be directed towards south. As the velocity \(\overrightarrow{\mathrm{υ}}\) of the electron is from west to east, the expression for the magnetic Lorentz force i.e. \(\overrightarrow{\mathrm{F}}\) m = – e \((\vec{v} \times \overrightarrow{\mathrm{B}})\) tells us that the magnetic field \(\overrightarrow{\mathrm{B}}\) should be applied along the vertical and in the downward direction.

Question 19.

An electron emitted by a heated cathode and accelerated through a potential difference of 2.0 kV, enters a region with uniform magnetic field of 0.15 T. Determine the trajectory of the electron if the field (a) is transverse to its initial velocity, (b) makes an angle of 30° with the initial velocity.

Answer:

Here, υ = potential difference = 2.0 kV = 2000 V,

B = uniform magnetic field = 0.15 T

Let e, m be the charge and mass of the electron.

Also let υ be the velocity acquired by the electron when accelerated under potential difference V.

(a) When the magnetic field is transverse to the initial velocity. The force on the electron due to transverse magnetic field =Be υ.

This force provides the necessary centrepetal force to the electron required for uniform circular motion.

So the trajectory of the electrons is circular having radius r (say) obtained as

(b) When the magnetic field makes an anele 30° with the initial velocity.

Here θ = 30°

∴ Component of v nonnal to \(\overrightarrow{\mathrm{B}}\) is υ_{2} = υ sin 30° \(\frac {1}{2}\)

or υ_{2} = \(\frac {8}{3}\) x 10^{7} x \(\frac {1}{2}\) = \(\frac {4}{3}\) x 10^{7} ms^{-1}

Due to υ_{2}, the force acting on the electron due to \(\overrightarrow{\mathrm{B}}\) will be providing the required centrepetal force, hence the electron will move along a circle.

is the component of velocity along \(\overrightarrow{\mathrm{B}}\) , so no force will act on the electron due to vy So the electron shall follow a straight path along \(\overrightarrow{\mathrm{B}}\) Due to these two component velocities, the electron will describe a helical path having radius, \(\overrightarrow{\mathrm{r}_{\mathrm{1}}}\) (say) given by

\(\overrightarrow{\mathrm{r}_{\mathrm{1}}}\) = \(\frac{\mathrm{m} v_{2}}{\mathrm{Be}}\)

(∴ υ_{2} = υ_{1} = component of \(\overrightarrow{\mathrm{υ}}\) normal to \(\overrightarrow{\mathrm{B}}\)

= \(\frac{9.0 \times 10^{-31} \times \frac{4}{3} \times 10^{7}}{0.15 \times 1.6 \times 10^{-19}}\)

= \(\frac {1}{2}\) mm = 0.5 mm

Question 20.

A magnetic field set up using Helmholtz coils (described in Exercise 4.16) is uniform in a small region and has a magnitude of 0.75 T. In the same region, a uniform electrostatic field is maintained in a direction normal to the common axis of the coils. A narrow beam of (single species) charged particles all accelerated through 15 kV enters this region in a direction perpendicular to both the axis of the coils and the electrostatic field. If the beam remains undeflected when the electrostatic field is 9.0 x 10+5 Vm”1, make a simple guess as to what the beam contains. Why is the answer not unique?

Answer:

Here, B = magnetic field =0.75 T

E = Electric field applied normal to the axis of coils in the same region of B = 9 x 10^{-5} Vm^{-1}

V = P.D. applied = 15 kV = 15000 V.

Let e, m be the charge and mass of the charged particle.

As the beam of charged particles remains undeflected, so the force due to magnetic and electric fields must be equal and opposite i.e.

Fm = Fe

or BeV = eE

or υ = \(\frac {E}{B}\) = \(\frac{9 \times 10^{+5}}{0.75}\) = 12 x 10^{+6} ms^{-1}

nature of particles in the beam.

Also \(\frac {1}{2}\) mυ^{2} = eV

or \(\frac {e}{m}\) = \(\frac{v^{2}}{2 \mathrm{~V}}\) = \(\frac{\left(12 \times 10^{6}\right)^{2}}{2 \times 15000}\)

= 4.8 x 10^{7} C kg^{-1}

Also, we know that \(\frac {e}{m}\) for proton is = 9.6 x 10^{-7} x C kg^{-1}. It follows m

that the charged particle under reference has the value of \(\frac {e}{m}\) half of that

for the proton, so its mass is clearly double the mass of proton. Thus the beam may be of deutrons.

The answer is not unique as the ratio of charge to mass i.e. 4.8 = 10^{7} C kg^{-1} may be satisfied by many other charged particles, such as He^{++} (\(\frac {2e}{2m}\)) and Li^{+++} (\(\frac {3e}{3m}\))

which have the same value of (\(\frac {e}{m}\))

Question 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 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 = 9.8m s^{2}A

Answer:

Here, l = length of rod = 0.45 m.

m = mass of rod = 60 g = 60 x 10^{-3} kg.

I = current passing through the rod 5.0 A

(a) Tire tension in the wire will be zero if the force on Jhe wire due to the magnetic field is equal and opposite to the weight of the rod and is in the vertically upward direction.

i.e BI l = mg

B = \(\frac{\mathrm{mg}}{\mathrm{I} l}=\frac{60 \times 10^{-3} \times 9.8}{5 \times 0.45}\) = 0.261 T

The force due to \(\overrightarrow{\mathrm{B}}\) will be upward if the direction of the field is horizontal and normal to the conductor as can be determined by applying Fleming’s left hand rule.

(b) When the direction of the current is reversed, then the force on the rod due to B will act in the direction of its weight i.e. vertically

downwards, so tension in the supporting wire will be given by,

T = BI l + mg

= 0.261 x 5 x 0.45 + 60 x 10^{-3} x 9.8

= 0.588+ 0.588 = 1.176 N.

i.e. T = mg + mg = 2 x weight of the rod.

Question 22.

The wires which connect the battery of an automoblie 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 of repulsive?

Answer:

Here, I_{1} = I_{2} = 300 A

r = 1.5 cm = 1.5 x 10^{2}m

Length of each wire, l = 70 cm = 70 x 10^{-2} m.

We know that the force per unit length of a current carrying long conductor is given by

F = \(\frac{\mu_{0}}{4 \pi} \times \frac{2 \mathrm{I}_{1} \mathrm{I}_{2}}{\mathrm{r}}\)

As the current in the two wires connecting the battery to the starting motor flows in opposite directions, so the force is repulsive.

Question 23.

A uniform magnetic field of 1.5 T exists in a cylindrical region of radius 10.0 cm, its direction parallel to the axis along east to west. A wire carrying current of 7.0 A in the north to south direction passes through this region. What is the magnitude and direction of the force on the wire if,

(a) the wire intersects the axis

(b) the wire is turned from N-S to northeast-northwest direction,

(c) the wire in the N-S direction is lowered from the axis by a distance of 6.0 cm?

Answer:

Here, B = uniform magnetic field = 1.5 T.

I = current in the wire = 7.0 A

(a) Length of the wire in a cylindrical magnetic field is equal to the diameter of the cylinderical region i.e.

Z = 2r = 2 x 10 = 20 cm = 20 x 10^{-2} m.

Angle between the current and the magneitc field, 0 = 90°

∴ Force on the wire is given by

F = BI l sin θ

= 1.5 x 7.0 x 20 x 10^{-2} x sin 90 = 2.1 N

The direction of this force is vertically downwards using Fleming’s left hand rule.

(b) When the wire is turned from N-S to north east – north west direction, then θ = 45° and length of the wire in the cylindrical region of the magneitc field is I_{1} (say) given by

I = I_{1} sin 45° or I_{1} = \(\frac{l}{\sin 45^{\circ}}\)

or I_{1} = \(\frac{l}{1 / \sqrt{2}}\) = \(\sqrt{2}\) l

If F_{1} be the force on the wire in this case, then

F_{1} = B I l_{1} sin 45°

= 1.5 x 7.0 x \(\sqrt{2}\) l x \(\frac{l}{1 / \sqrt{2}}\)

= 10.5 x 0.2 = 2.1 N = 2.1 N

The force acts vertically downwards.

(c) When the wire is lowered by-6.0 cm i.e. taken at position CD as shown in the figure here, then the length of the wire in magnetic field is given by

l_{2} = 2x

where x is given by

x.x = 4 x (10 + 6) = 64

or x^{2} = 64

x.x = 8cm.

l_{2} = 2 x 8

= 16 cm

= 16 x 10^{-2}

Let F_{2} be the force on the wire in this case,

F_{2} = BI l_{2}

= 1.5 x 7x 10 x 10^{-2} = 1.68 N

and it acts vertically downwards.

Question 24.

A uniform magentic field of 3000 G is established along the positive z-direction. A reactangular 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 Fig. What is the force on each case? Which case corresponds to stable equlibrium ?

answer:

Here, B = unifrom megnetic feild gasus along z-axis = 3000 x 10^{-4} T = 0.3 T

l = length of rectangular loop

= 10 cm = 0.1 m

b = breath of rectangular loop

= 5 cm = 0.05 m

∴ A = area of rectangular loop

= l x b = 10 x 5 = 50 cm^{2} = 50 x 10^{-4} m^{2}

Torque on the loop is given by

τ = IAB cos θ

When θ is the angle between the plane of the loop and the direction of magnetic field.

(a) Here θ = 0

τ = 0.3 x 12 x 50 x 10^{-4} x cos 0

= 1.8 x 10^{2} Nm

and acts along y’- axis i.e. – y direction

(b) θ = 0

τ = 0.3 x 12 x 50 x 10 x cos 0

= 1.8 x 10^{2} Nm

and acts along y’-axis or y-axis

(c) θ = 0

∴ τ = 0.3 x 12 x 50 x 10 x cos 0

= 1.8 x 10^{2} Nm

and acts along – x direction or x’- axis

(d) Here θ = 0

∴ τ = 1.8 x 10^{-2}

(e) Here θ = 90^{0}, so cos 90 = 0

∴ τ = BIA cos 90 = 0

(f) θ = 90 τ = 0

Aliter:

\(\overrightarrow{\mathrm{τ}}\) = \((\overrightarrow{\mathrm{A}}) \times \overrightarrow{\mathrm{B}}\)

IA = 50 x 10^{4} x 12 = 0.06 Am^{+2}

(a) Here, I \(\overrightarrow{\mathrm{A}}\) = 0.06 \(\hat{\mathrm{i}}\) ,Am^{2}, \(\overrightarrow{\mathrm{B}}\) = 0.3 \(\hat{\mathrm{k}}\) T

same as in case (a)

(c) I \(\overrightarrow{\mathrm{A}}\) = 0.06 \(\hat{\mathrm{i}}\) ,Am^{2}, \(\overrightarrow{\mathrm{B}}\) = 0.3 \(\hat{\mathrm{k}}\) T

∴ \(\overrightarrow{\mathrm{τ}}\) = \((\overrightarrow{\mathrm{A}}) \times \overrightarrow{\mathrm{B}}\) = 0.06\(\hat{\mathrm{j}}\) x 0.3 \(\hat{\mathrm{k}}\)

= -1.8 x 10^{-2} \(\hat{\mathrm{j}}\) Nm

∴ τ = 1.8 x 10^{-2} Nm and acts along – x direction

(d) Here \(\overrightarrow{\mathrm{A}}\) = \(\overrightarrow{\mathrm{l}}\) x \(\overrightarrow{\mathrm{b}}\)

= 0.1 \(\hat{\mathrm{k}}\) x (0.05 cos 300 \(\hat{\mathrm{j}}\) + 0.05 sin 30\(\hat{\mathrm{i}}\))

= 5 x 10^{-3} \(\frac{\sqrt{3}}{2}\) \(\hat{\mathrm{k}}\) x \(\hat{\mathrm{j}}\) + 5 x 10^{-3} x \(\frac { 1 }{ 2 }\)\(\hat{\mathrm{k}}\) x \(\hat{\mathrm{i}}\)

= 2.5 x 10^{-3} \((-\sqrt{3} \hat{i}+\hat{i})\)

\(\overrightarrow{\mathrm{τ}}\) = \((\overrightarrow{\mathrm{A}}) \times \overrightarrow{\mathrm{B}}\)

= 12 x 2.5 x 10^{-3} \((-\sqrt{3} \hat{i}+\hat{j})\) x 0.3 \(\hat{\mathbf{k}}\)

= 9 x 10^{-3} \((-\sqrt{3} \hat{i} \times \hat{k}+\hat{j} \times \vec{k})\)

= 9 x 10^{-3}

= 9 x 10^{-3}\((\hat{i}+\sqrt{3} \hat{j})\) Nm.

∴ magnitude of torque is given by \(\overrightarrow{\mathrm{τ}}\) is given by

\(\overrightarrow{\mathrm{τ}}\) = \(\sqrt{\left(9 \times 10^{-3}\right)^{2}+\left(9 \times 10^{-3} \times \sqrt{3}\right)^{2}}\)

= \(\sqrt{\left(9 \times 10^{-3}\right)^{2} \times(1+3)}\)

= \(\sqrt{\left(9 \times 10^{-3} \times 2\right)^{2}}\)

= 18 x 10^{-3} = 1.8 x 10^{-2} Nm

Direction of \(\overrightarrow{\mathrm{τ}}\) :

Let (1 be the angle made by \(\overrightarrow{\mathrm{τ}}\) with + ve direction of x-axis

using \(\overrightarrow{\mathrm{A}}\) . \(\overrightarrow{\mathrm{B}}\) = AB cos θ, we get

\(\overrightarrow{\mathrm{τ}}\) . \(\hat{\mathrm{i}}\) = (\(\overrightarrow{\mathrm{τ}}\)) . (\(\hat{\mathrm{i}}\) ) cos β

= τ . 1 . cos β

(e) Here, \(\mathrm{I} \overrightarrow{\mathrm{A}}\) = 0.06 \(\hat{\mathrm{k}}\) Am^{2}, \(\overrightarrow{\mathrm{B}}\) = 0.3 \(\hat{\mathrm{k}}\) T

\(\overrightarrow{\mathrm{τ}}\) = 0.06 \(\hat{\mathrm{k}}\) x 0.3 \(\hat{\mathrm{k}}\) = 0

(f) Here, \(\mathrm{I} \overrightarrow{\mathrm{A}}\) = -0.06 \(\hat{\mathrm{k}}\) Am^{2}, \(\mathrm{I} \overrightarrow{\mathrm{B}}\) = 0.3 \(\hat{\mathrm{k}}\) T

∴ \(\overrightarrow{\mathrm{τ}}\) = -0.06 x 0.3 \(\hat{\mathrm{k}}\) x \(\hat{\mathrm{k}}\) = 0

Net force on a planar loop in a magnetic field is always zero, so force is zero in each case.

Case (e) corresponds to stable equilibrium as I\(\overrightarrow{\mathrm{A}}\) is aligned with \(\overrightarrow{\mathrm{B}}\) while \(\overrightarrow{\mathrm{A}}\) corresponds to unstable equilibrium as IA is antiparallel to \(\overrightarrow{\mathrm{B}}\)

Question 25.

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 0.5 A, what is tjre

(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} m^{2}, and the free electron density in copper is given to be about 10^{29} m^{-3},)

Answer:

(a) Here, n = number of turns in the circular coil = 20

r = radius of the circular coil = 10 cm = 10 x 10^{-2} m.

B = uniform magnetic field = 0.10 T

θ = angle between \(\overrightarrow{\mathrm{B}}\) and normal to the plane of coil = 0°

I = current in the coil = 5.0 A

A = Area of the coil = πr^{2}

= \(\frac{22}{7} \times(0.10)^{2}\) = m^{2}

τ = torque on the coil = ?

(a) Using formula,

τ = n B I A sin θ

We get, τ = 20 x 5 x 0.10 x \(\frac{22}{7}\) x (0.10)^{2} x sin θ°

= 0

(b) Total force on a planar current loop in a magnetic field is always zero.

(c) Force on an electron is given by

F = B e υd = Be \(\left(\frac{\mathrm{I}}{\mathrm{neA}^{\prime}}\right)=\frac{\mathrm{BI}}{\mathrm{nA}}\)

(∴ I = ne A vd)

Here, n = 10^{29} m^{-3}

= free electron density in copper

A’ = area of cross-section of copper wire = 10^{-5} m^{2}

F = \(\frac{0.10 \times 5}{10^{29} \times 10^{-5}}\)

= 5 x 10^{-25} N

Question 26.

A solenoid 60 cm long and of radius 4.0 cm has 3 layers of windings of 300 turns each. A 2.0 cm long wire of mass 2.5 g lies inside the solenoid (near its centre) normal to its axis; both the wire and the axis of the solenoid are in the horizontal plane. The wire is connected through two leads parallel to the axis of the solenoid to an external battery which supplies a current of 6.0 A in the wire. What value of current (with appropriate sense of circulation) in the windings of the solenoid can support the weight of the wire? g = 9.8 m s^{-2}

Answer:

Let I be the current passed through the solenoid.

If B = the magnetic field set up at the centre of the solenoid, then

B = μ_{0} nl ……..(i)

This field acts normally to the length of the current carrying wire, thus the wire experiences a force F given by

F = BI’ l …… (ii)

When I’ = Current flowing through the wire = 6.0 A. and l = length of the wire placed inside the solenoid = 2 cm = 0.02 m

Here, N = total no. of turns = no. of layers x no. of turns in each layer .

= 3 x 300 = 900

L = length of solenoid

= 60 cm = 0.6 m

r = radius of solenoid

4 cm = 0.04 m

n = \(\frac {N}{L}\) = 1500 turns m^{-1}

∴ From (1) B = 4π x 10^{-7} x 1500 x I

6π x 10^{-4} I Tesla.

6π x 10^{-4} x 6 x 0.02

2.262 x 10^{-4} I N

The direction of current through the solenoid and hence the magnetic field are understood to be in such a direction that the force F acts upwards and the wrightof wire will be supported only if F is equal and opposite to the weight of the wire i.e.

F = mg

or 2.262 x 10^{-4} I = mg

m = 2.5 x 10^{-3} kg

g = 9.5 ms^{-2}

∴ 2.262 x 10^{-4} I = 2.5 x 10^{-3} x 9.8

0r I = \(\frac{2.5 \times 10^{-3} \times 9.8}{2.262 \times 10^{-4}}\)

= 108.3 A

Question 27.

A galvanometer coil has a resistance of 12 Ω and the meter shows full scale deflection for a current of 3 mA. How will you convert the meter into a voltmeter of range 0 to 18 V?

Answer:

Here, G = galvanometer resistance = 12 Ω

I_{g} = Galvanometer full scale deflection current

= 3mA = 3 x 10^{-3} A

V = 18V

R = resistance to be connected in series for conversion of galvanometer into voltmeter of 0 -18 V range = ?

Using the relation

V = – I_{g} (R + G), we get

R =\(\frac{\mathrm{V}}{\mathrm{I}_{\mathrm{g}}}\) – G

= \(\frac{18}{3 \times 10^{-3}}\) – 12

= 6000 – 12 = 5988 Ω

Question 28.

A galvanometer coil has a resistance of 12 Ω and the meter shows full scale deflection for a current of 4 mA. How will you convert the meter into a voltmeter of range 0 to 6 V?

Answer:

Here, galvanometer resistance, G = 15 Ω

I_{g} = current following in the galvanometer corresponding to which it gives full scale defection = 4 mA =4x 10^{-3} A

I = total current to be measured by an ammeter = 6A.

S = Shunt resistance to be connected in parallel to the galvanometer = ?

Using the relation,

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