WBCHSE Class 12 Physics Electromagnetism Short Question And Answers

Magnetic Effect Of Current And Magnetism

Electromagnetism Short Question And Answers

Question 1. Two moving coil meters M₁ and M₂ have the following particulars

(The spring constants are identical for the two meters.) Determine the ratio of

1. Current sensitivity and
2. Voltage sensitivity of M₂ and Mx.

Answer:

Current sensitivity = \(\frac{N B A}{c}\) [c = restoring torque for unit deflection]

∴ \(\frac{\text { sensitivity of } M_2}{\text { sensitivity of } M_1}=\frac{N_2 B_2 A_2}{N_1 B_1 A_1} \quad\left[∵ c_1=c_2\right]\)

⇒ \(\frac{42 \times 0.50 \times 1.8 \times 10^{-3}}{30 \times 0.25 \times 3.6 \times 10^{-3}}\)

= 1.4

2. \(\text { Voltage sensitivity }=\frac{N B A}{c R}=\frac{\text { current sensitivity }}{R}\)

∴ \(\begin{aligned}
& \text { voltage sensitivity } \\
& \frac{\text { of } M_2}{\text { voltage sensitivity }}=1.4 \times \frac{R_1}{R_2}=1.4 \times \frac{10}{14}=1 \\
& \text { of } M_1 \\
&
\end{aligned}\)

WBBSE Class 12 Electromagnetism Short Q&A

Question 2.

1. A circular coil of 30 turns and a radius of 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 60° with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.
2. 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 unaltered.)

Answer:

1. The torque acting on the coil,

⇒ \(\tau=N B I A \sin \theta\)

=30 x 1 x 6 x π x (8x 10-2)2sin 60°

= 3.13 N.m

∴ If a torque of 3.13 N m is applied in the opposite direction, there will be no deflection of the coil.

2. The torque does not depend on the shape of the coil if the area remains the same. So the answer will not change

Question 3.Two concentric circular coils X and Y of radii 16 cm and 10 cm, respectively, lie in the same vertical plane oriented along 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 in Y is clockwise, 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 center.
Answer:

⇒ \(B=\frac{\mu_0 n l}{2 a}\) [ n = no. of turns, I = current a = radius of the coil]

∴ For coil X,

⇒ \(B_X=\frac{4 \pi \times 10^{-7} \times 20 \times 16}{2 \times 16 \times 10^{-2}}\)

= 4π x 10^-4 T (towards east)

For coil Y,

⇒ \(B_Y=\frac{4 \pi \times 10^{-7} \times 25 \times 18}{2 \times 10 \times 10^{-2}}\)

= 9π x 10^-4 T (towards west)

⇒ \(\text { Net field }=B_Y-B_X=9 \pi \times 10^{-4}-4 \pi \times 10^{-4}\)

= 1.6 x 10^-3 T (towards west)

Short Answer Questions on Faraday’s Law

Question 4. A uniform magnetic field of 100 G (1G = 10^-4 T) exists in a region of length about 10 cm and an area of cross-section of about 10^-3 m². The maximum current carrying capacity of a given coil of wire is 15A and the number of turns per unit length that can be wound around a core is at most 1000 turns m^-1. How would you utilize the coil to design a solenoid for the required purpose? Assume the core is not ferromagnetic.
Answer:

For the maximum number of turns the current through the solenoid is given by,

⇒ \(I=\frac{B l}{\mu_0 N}=\frac{B}{\mu_0\left(\frac{N}{l}\right)}\left[∵ B=\mu_0 \cdot \frac{N}{l} \cdot I\right]\)

⇒ \(\frac{100 \times 10^{-4}}{4 \pi \times 10^7 \times 100} \mathrm{~A}=7.96 \mathrm{~A} \approx 8 \mathrm{~A}\)

The length of the solenoid should be much more than its diameter.

For a current of 15 A to flow through the solenoid, the number of turns per unit length has to be decreased [as \(I \propto \frac{l}{N} \propto \frac{1}{N / l}\)]

∴ \(\frac{I^{\prime}}{I}=\frac{N / l}{N^{\prime} / l^{\prime}} \quad \text { or, } \frac{15}{8}=\frac{1000}{N^{\prime} / l^{\prime}}\)

∴ \(\frac{N^{\prime}}{l^{\prime}}=\frac{1000 \times 8}{15}=533 \approx 550 \text { turns } \cdot \mathrm{m}^{-1}\)

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

⇒ \(B=\frac{\mu_0 I R^2 N^{\prime}}{2\left(x^2+R^2\right)^{3 / 2}}\)

Question 5. 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 compared to R and is given by \(B=0.72 \mu_0 N I / R\), approximately. (Such an arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.)
Answer:

The magnetic field in a small region of length 2d about the midpoint of the space between the two coils

⇒ \(B=\frac{\mu_0 N I R^2}{2\left[\left(\frac{R}{2}+d\right)^2+R^2\right]^{3 / 2}}+\frac{\mu_0 N I R^2}{2\left[\left(\frac{R}{2}-d\right)^2+R^2\right]^{3 / 2}}\)

⇒ \(\frac{\mu_0 N I R^2}{2}\left\{\left[\frac{5 R^2}{4}\left(1+\frac{4 d}{5 R}\right)\right]^{-\frac{3}{2}}+\left[\frac{5 R^2}{4}\left(1-\frac{4 d}{5 R}\right)\right]^{-\frac{3}{2}}\right\}\)

⇒ \(=0.72 \frac{\mu_0 N I}{R}\)

Common Short Questions on Lenz’s Law

Question 6. 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 in the environment?
Answer:

Yes. The direction of the velocity of a charged particle can change in a magnetic field, but the magnitude of velocity remains the same.

Question 7. An electron traveling west to east enters a chamber having a uniform electrostatic field in the 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:

The negatively charged electron tends to move towards the north due to the electric field. If an equal magnetic force acts in the south direction then the electron will not be deflected. Using the hand rule and considering the conventional direction of flow of charge from east to west we get the direction of magnetic field along the vertically download direction.

Question 8. An electron emitted by a heated; cathode and accelerated through a potential difference of 2.0 kV, enters a region of 1 uniform magnetic field of 0.15 T. Determine the trajectory of the electron if the field

1. Is transverse to its initial velocity,
2. Makes an angle of 30° with the initial velocity.

Answer:

1. The trajectory of the electron will be a circle of radius r with the magnetic field as its axis. The radius r is given by,

⇒ \(r=\frac{m v}{B e}\)

⇒ \(\text { Also, } e V=\frac{1}{2} m v^2\)

∴ \(r=\frac{1}{B}\left(\frac{2 m V}{e}\right)^{\frac{1}{2}}=\frac{1}{0.15} \sqrt{\frac{2 \times 9 \times 10^{-31} \times 2 \times 10^3}{1.6 \times 10^{-19}}}\)

= 1mm

2. The electron will move along the direction of the field with velocity, v’ = ysin30°, and radius r’ with the magnetic field as the axis.

⇒ \(r^{\prime}=\frac{m v^{\prime}}{B e}=\frac{\sin 30^{\circ}}{B} \sqrt{\frac{2 m V}{e}}\)

⇒ \(=\frac{\frac{1}{2}}{0.15} \sqrt{\frac{2 \times 9 \times 10^{-31} \times 2 \times 10^3}{1.6 \times 10^{-19}}}\)

= 0.5 x 10^-3 m²

The electron will follow a helical path. Its pitch is,

⇒ \(2 \pi r^{\prime} \cot \theta=2 \times 3.14 \times\left(0.5 \times 10^{-3}\right) \times \sqrt{3}\)

= 5.44 X 10^-3 m

= 5.44 mm

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Practice Short Questions on Induced EMF

Question 9. A magnetic field, set up using Helmholtz coils, 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 15kV, 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 10⁵ V.m^-1, make a simple guess as to what the beam contains. Why is the answer not unique?
Answer:

qE = Bqv

∴ \(v=\frac{E}{B}=\frac{9 \times 10^5}{0.75}=1.2 \times 10^6 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

⇒ \(\frac{1}{2} m v^2=q V\)

∴ \(\frac{q}{m}=\frac{1}{2} \frac{v^2}{V}=\frac{1}{2} \times \frac{\left(1.2 \times 10^6\right)^2}{15 \times 10^3}=4.8 \times 10^7 \mathrm{C} \cdot \mathrm{kg}^{-1}\)

He⁺⁺ ‘, Li⁺⁺⁺ ₁H² ion – for all these particles

⇒ \(\frac{q}{m}=4.8 \times 10^7 \mathrm{C} \cdot \mathrm{kg}^{-1}\)

∴ The charged particle may be any of these.

Question 10. A straight horizontal conducting rod of length 0.45 m and 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.

1. What magnetic field should be set up normally to the conductor in order that the tension in the wires is zero?
2. What will be the total tension in the wires if the direction of the current is reversed keeping the magnetic field the same as before? (Ignore the mass of the wires) 9.8 ms^-2

Answer:

1. To keep the tension in the wire zero, the upward force exerted by the magnetic field should be neutralized by the downward force exerted by gravity (weight of the rod).

∴ BIl = mg

∴ \(B=\frac{m g}{I l}=\frac{60 \times 10^{-3} \times 9.8}{5 \times 0.45}=0.26 \mathrm{~T}\)

2. If the direction of the current is reversed, the tension in the wire will be doubled.

∴ T = 2 x weight of the rod

= 2 x 60 x 10^-3 x 9.8 N

= 1.18 N

Question 11. A uniform magnetic field of 3θ 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? Which case corresponds to stable equilibrium?

Answer:

Using \(\tau\) = IAsinθ

1. \(\tau\) = 12 x (10 x 5 x 10^-4) x (3θ X 10^-4)

= 1.8 x 10^-2 N.m^-1 , along y-axis

2. \(\tau\) = 1.8 x 10^-2 N m^-1 , along y-axis

3. \(\tau\) = 1.8 x 10^-2 N.m^-1, along -X axis

4. \(\tau\) = 1.8 x 10^-2 N.m^-1 , along a direction making 240° with +X axis

5. \(\tau\) = 0, stable equilibrium, because if this loop is slightly rotated, the torque generated tends to bring the loop back to its initial alignment.

6. \(\tau\) = 0, unstable equilibrium, because if the loop Is slightly rotated, the torque generated tends to rotate the loop farther.

Question 12. A 60 cm long solenoid 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 center) 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:

⇒ \(F=B I l \text { and } B=\mu_0 \cdot \frac{N}{l^{\prime}} \cdot I^{\prime}\)

∴ \(F=\mu_0 \cdot \frac{N}{l’} \cdot l’ \cdot l \cdot l\) [where I’m required current, I’ length of the solenoid, I current from the battery, I length of the wire.

This force will neutralize the gravitational force on the wire.

∴ \(\mu_0 N \cdot \frac{l}{l} \cdot I^{\prime} \cdot I=m g\)

or, \(I^{\prime}=\frac{m g}{\mu_0 N l \cdot l}\)

⇒ \(\frac{2.5 \times 10^{-3} \times 9.8 \times 60 \times 10^{-2}}{4 \pi \times 10^{-7} \times 3 \times 300 \times 2.0 \times 10^{-2} \times 6.0}\)

=108.3A

Question 13. Two long and parallel straight wires A and B carrying currents of 0.0 A and 5.0 A In the same direction are separated by a distance of 4.0 cm listen to the force on a 10 cm section of wire A.
Answer:

We know that, \(\frac{F}{l}=\frac{\mu_0 I_1 I_2}{2 \pi r} \quad\)

or, \(F=\frac{\mu_0 I_1 I_2}{2 \pi r} \cdot l\)

∴ \(F=\frac{4 \pi \times 10^{-7} \times 8 \times 5}{2 \pi \times 4 \times 10^{-2}} \times 10 \times 10^{-2}=2 \times 10^{-5} \mathrm{~N}\)

(attractive force Is towards A to B along the perpendicular)

Question 14. A uniform magnetic 1.5 T exists in a cylindrical region of radius 10.0 cm. Its direction is parallel to the axis along east to west. A wire carrying a 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 lf,

1. The wire intersects the axis
2. The wire Is turned from N-S to the northeast-northwest direction
3. The wire In the N-S direction Is lowered from the axis by a distance of 6.0 cm

Answer:

1. F₁ = IBl sin90° = 1.5 X 7 X 0.2 x 1 = 2.1 N (directed vertically downwards)

2. If l₁, be the effective length of the wire Inside the magnetic field,

F₂ = IBl₁ sin45° = IBl [since, l₁ sin45° = l]

= 1.57 x 7 x 0.2

= 2.1 N (directed vertically, downwards)

3. When wire 16 is lowered by 6 cm from the axis, then the effective length of the wire inside the magnetic field is 2α.

a = 8 cm

F = IB.2a

= 1.5 x 7 x 0.16N

= 1.68 N (directed vertically downwards)

Important Definitions in Electromagnetism

Question 15. A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil i$ 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:

Torque

⇒ \(\tau\) = NIABsinθ

= 20 x 12 x (10 x 10^-2)² x 0.8sin30°

= 0.96 N.m [Here, N= number of turns, I=current, A – area of the coil, B = magnetic field]

Question 16. 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

1. Outside the toroid
2. Inside the core of the toroid
3. In the empty space surrounded by the toroid

Answer:

1. Zero

2. \(B=\mu_0 \frac{N I}{2 \pi r}=\frac{4 \pi \times 10^{-7} \times 3500 \times 11}{2 \pi \times 25.5 \times 10^{-2}}\) \(\left[\text { Here } r=\frac{25+26}{2}=25.5 \mathrm{~cm}\right]\)

= 3.02 x 10^-2 T

3. Zero

Question 17. A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8cm. If the current carried is 8.0 A, estimate the magnitude of B inside the solenoid near its center.
Answer:

⇒ \(B=\frac{\mu_0 \cdot N I}{l}=\frac{4 \pi \times 10^{-7} \times 400 \times 8 \times 5}{80 \times 10^{-2}}\)

= 8π x 10^-3T

Question 18. A electron is moving with a velocity \(\vec{v}=(\hat{i}+2 \hat{j}) \mathrm{m} \cdot \mathrm{s}^{-1}\) in the magnetic field \(\vec{B}=(2 \hat{i}+2 \hat{j}) \mathrm{Wb} \cdot \mathrm{m}^{-2}\). Determine the magnitude and direction of the force acting on the electron. The charge of an electron is -1.6 x 10^-19 C.
Answer:

The force acting on the electron,

⇒ \(\vec{F}=q(\vec{v} \times \vec{B})=-1.6 \times 10^{-19}[(\hat{i}+2 \hat{j}) \times(2 \hat{i}+2 \hat{j})]\)

⇒ \(-1.6 \times 10^{-19} \times[-2 \hat{k}]=+3.2 \times 10^{-19} \hat{k} \mathrm{~N}\)

So the magnitude of the force = 3.2 x 10^-19N and the direction is along the positive z-axis.

Question 19. Which physical quantity has the unit Wb/m²? Is it a scalar or a vector quantity?
Answer:

Magnetic field or magnetic induction has the unit Wb/m². It is a vector quantity.

Question 20. Write down the equation of Lorentz force acting on a moving charged particle.
Answer:

Lorentz force, \(\vec{F}=q(\vec{E}+\vec{v} \times \vec{B})\);

where \(\vec{E}\) is the electric field,

⇒ \(\vec{B}\) is the magnetic field,

q is the charge of the particle

and \(\vec{v}\) is the velocity ofthe particle.

Question 21.

1. α – particle
2. β – particle is both projected with the same velocity v perpendicularly to the magnetic field B. Which particle will experience greater force?

Answer:

The value of the charge of the a -particle (+2e) is more than that of the β-particle (-e).

Hence, the α-particle will experience greater force

Question 22. In a compact coil of 50 turns, the current strength is 10 A and the radius of the coil is 25 x 10^-2 m. Find the magnitude of the magnetic field at its center.
Answer:

Number of turns, N = 50; radius of the coil, r = 25 x 10^-2 m; current strength, I = 10 A.

∴ The magnetic field at the center of the coil,

⇒ \(B=\frac{\mu_0 N I}{2 r}=\frac{4 \pi \times 10^{-7} \times 50 \times 10}{2 \times 25 \times 10^{-2}}\)

= 1.256 X 10^-3T

Real-Life Scenarios in Electromagnetism

Question 23. Two long parallel straight wires P and Q separated by a distance of 5 cm in air carry currents of 4A and 2A respectively in the same direction. Fruk the magnitude of the force acting per cm of the wire P and indicate the direction of the force.
Answer:

Distance between the two wires, \(r=5 \mathrm{~cm}=\frac{5}{100} \mathrm{~m}\)

Force acting per metre, \(F=\frac{\mu_0}{4 \pi} \cdot \frac{2 I_1 I_2}{r}=\frac{4 \pi \times 10^{-7}}{4 \pi} \times \frac{2 \times 4 \times 2}{\frac{5}{100}}\)

⇒ \(10^{-7} \times \frac{16}{5} \times 100=3.2 \times 10^{-5} \mathrm{~N} \cdot \mathrm{m}^{-1}\)

∴ Force acting per cm = \(=\left(3.2 \times 10^{-5}\right) \times \frac{1}{100}\)

= 0.032 dyn.cm^-1

Since the wires are carrying currents in the same direction, the force acting between the wires is attractive, Hence the force on each wire is directed towards the other wire and perpendicular to them.

Question 24. Write the expression for the force \(\vec{F}\) acting on a particle of charge q moving with a velocity \(\vec{v}\) in the presence of both electric fields \(\vec{E}\) and magnetic field \(\vec{B}\). Obtain the condition under which the particle moves undeflected through the fields.
Answer:

1st Part: Force,

⇒ \(\vec{F}=\vec{F}_e+\vec{F}_m=q \vec{E}+q \vec{v} \times \vec{B}=q(\vec{E}+\vec{\nu} \times \vec{B})\)

2nd Part: The required condition is, either \(\vec{F}\) = 0; i.e., no resultant force acts on the particle. In that case,

⇒ \(q \vec{E}=-q \vec{v} \times \vec{B}\)

or, \(\vec{E}\) and \(\vec{B}\) are both along the direction of velocity of the particle; then \(\vec{\nu} \times \vec{B}=0\), and the force \(q \vec{E}\) along the direction of motion produces a constant acceleration without any deflection.

Question 25. A wire AB is carrying a steady current of 12 A and is lying on a table. Another wire CD carrying 5 A is held directly above AB at a height of 1 mm. Find the mass per unit length of the wire CD so that it remains suspended at its position when left free. Give the direction of the current flowing in CD with respect to that in AB. [Thke the value of g = 10 m.s^-2 ]
Answer:

We take the unit length of each of the wires.

Upward force on CD,

⇒ \(F=\frac{\mu_0}{4 \pi} \frac{2 I_1 I_2}{d}=10^{-7} \times \frac{2 \times 12 \times 5}{10^{-3}}=12 \times 10^{-3} \mathrm{~N}\)

Downward force.on CD,

F’ = mg = 10m N [ m = mass of unit length of the wire CD ]

Under the given conditions,

F = F’

or, 12 x 10^-3 = 10m

or, m = 1.2 x 10^-3 kg

The currents in AB and CD are in opposite directions, so that CD is subjected to an upward repulsive force F.

Question 26. 

1. Using Biot-Savart’s law, derive the expression for the magnetic field in vector form at a point on the axis of a circular current loop.
2. What does a toroid consist of? Find out the expression for the magnetic field inside a toroid for N turns of the coil having an average radius r and carrying a current I. Show that the magnetic field in the open space inside and outside the toroid is zero.

Answer:

1. From the text, we get the expression,

⇒ \(B=\frac{\mu_0 N I}{2} \cdot \frac{r^2}{\left(r^2+x^2\right)^{3 / 2}}\)

Here, the center of the coil has been taken as the origin and the direction of the axis as the x-axis.

∴ \(\vec{B}=\frac{\mu_0 N I}{2} \cdot \frac{r^2}{\left(r^2+x^2\right)^{3 / 2}} \hat{i}\)

2. Circumference of the toroid = 2πr.

Number of turns per unit length, \(n=\frac{N}{2 \pi r}\)

Magnetic field inside the toroid, \(B=\mu_0 n I=\frac{\mu_0 N I}{2 \pi r}\)

The currents in every pair of diametrically opposite points of the toroid is equal and opposite. So, if we take Ampere’s loops in the open space inside and outside the toroid, the current enclosed by the loops respectively, 0-0 = 0 and Nl-NI = 0

So, from Ampere’s circuital law, \(\oint \vec{B} \cdot d \vec{l}=\mu_0 I\) we have B = 0 in both cases.

Examples of Electromagnetic Applications

Question 27. What do you mean by the current sensitivity of a galvanometer? Write its SI unit.
Answer:

1st Part: The basic definition of the current sensitivity of a galvanometer is, \(S_i=\frac{d \theta}{d I}, \theta\), being the angular deflection of the coil.

2nd Part: As θ is dimensionless, the unit of S_i is A^-1.

Question An a -particle and a proton are released from the center of a cyclotron and made to accelerate.

1. Can both be accelerated at the same cyclotron frequency?
2. When they are accelerated, in turn, which of the two will have a higher velocity at the exit slit of the dees?

Answer:

Cyclotron frequency, \(n=\frac{1}{2 \pi}\left(\frac{q}{m}\right) B\)

1. For a proton, \(\frac{q}{m}=\frac{e}{m_p}\)

for an α-particle, \(\frac{q}{m} \approx \frac{2 e}{4 m_p}=\frac{1}{2} \frac{e}{m_p} .\)

⇒ \(\text { As } n \propto \frac{q}{m}\) they cannot be accelerated at the same
cyclotron frequency.

2. Velocity at the exit slit, \(v_0=\frac{q B R}{m} \text {, i.e., } v_0 \propto \frac{q}{m}\)

As \(\frac{q}{m}\) is higher for protons, it will hate a higher
velocity at the exit slit.

Question 28. Asha’s uncle was advised by his doctor to have an MRI (magnetic resonance imaging) scan of his brain. Her uncle felt that it was too expensive and wanted to postpone it. When Asha learned about this, she took the help of her family and when she approached the doctor, he also offered a substantial discount. She thus convinced her uncle to undergo the test to enable the doctor to know the condition of his brain. The resulting information greatly helped his doctor to treat him properly. Based on the above paragraph, answer the following questions.

1. What according to you are the values displayed by Asha, her family, and the doctor?
2. What in your view could be the reason for MRI tests to be so expensive?
3. Assuming that the Mill test was performed using a magnetic field of 0.1 T, and the maximum and minimum values of the force that the magnetic field could exert on a proton (Charge = 1.6 x 10^-19 C ) that was moving with a speed of 10⁴ m/s.

Answer:

1. Asha, her family, and the doctor were sympathetic, caring, socially aware, and helpful.

2. MRI tests are expensive due to various factors. Firstly, the MRI machines are very expensive and the rooms where the scans are performed cost a fortune. To this the salaries of the doctors, other medical staff, hospital charges, and maintenance charges add up, resulting in a huge amount.

3. We know, F = qvBsinθ

= 1.6 x 10^-19 x 104 x 0.1 x sinθ

We get the minimum value of force when θ = 0°.

∴ F_min = 0

We get the maximum value of force when θ = 90°

∴ \(F_{\max }=1.6 \times 10^{-19} \times 10^4 \times 0.1=1.6 \times 10^{-16} \mathrm{~N}\)

Question 29. What can be the cause of the helical motion of a charged particle?
Answer:

A particle moves along a curved path if it moves in the presence of a magnetic field.

Now, when there is an angular projection, i.e. when the velocity of the particle is at an angle with respect to the magnetic field, then the particle will move in a helical path.

The velocity component (vsinθ) perpendicular to the field will rotate the particle in a circular path. But, the component (vcosθ) along this field will move the particle in a straight line path. So, the cumulative motion of the particle is a helix.

Question 30. Two identical circular coils, P and Q each of radius R, carrying currents 1 A and √3 A respectively, are placed concentrically and perpendicular to each other lying in the XY and YZ planes. Find the magnitude and direction of the net magnetic field at the center of the coils
Answer:

Field due to current in coil P is \(\vec{B}_1=\frac{\mu_0 I_1}{2 R} \cdot \hat{k}\)

Current in coil Q is \(\vec{B}_2=\frac{\mu_0 I_2}{2 R} \cdot \hat{i}\)

Net field \(\vec{B}=\vec{B}_1+\vec{B}_2\)

⇒ \(\vec{B}=\left(\frac{\mu_0 I_1}{2 R}\right) \hat{k}+\left(\frac{\mu_0 I_2}{2 R}\right) \hat{i}\)

⇒ \(=\left(\frac{\mu_0}{2 R}\right) \hat{k}+\left(\frac{\sqrt{3} \mu_0}{2 R}\right) \hat{i}\left(∵ I_1=1 \mathrm{~A} ; I_2=\sqrt{3} \mathrm{~A}\right)\)

⇒ \(|\vec{B}|=\sqrt{\left(\frac{\mu_0}{2 R}\right)^2+\left(\frac{\sqrt{3} \mu_0}{2 R}\right)^2}=\frac{\mu_0}{2 R} \sqrt{1+3}=\frac{\mu_0}{2 R} \times 2\)

∴ \(|\vec{B}|=\frac{\mu_0}{R}\)

The resultant magnetic field is directed in the xz plane.

Question 31. Two identical loops P and Q each of radius 5 cm are lying in perpendicular planes such that they have a common center as shown in the figure. Find the magnitude and direction of the net magnetic field at the common center of the two coils, if they carry currents equal to 3 A and 4 A respectively.

Answer:

Magnetic field induction due to the vertical loop at the center O is,

⇒ \(B_1=\frac{\mu_0 I_1}{2 R}=\frac{\mu_0}{10^{-1}} . \quad(∵ R=5 \mathrm{~cm})\)

Magnetic field induction due to the horizontal loop at the center O is,

⇒ \(B_2=\frac{\mu_0 I_2}{2 R}=\frac{3 \mu_0}{10^{-1}}\)

∵ B₁ and B₂ are perpendicular to each other, therefore the resultant magnetic field induction at the center O is,

⇒ \(B_{\text {net }}=\sqrt{B_1^2+B_2^2}\)

⇒ \(\sqrt{\left(\frac{4 \mu_0}{10^{-1}}\right)^2+\left(\frac{3 \mu_0}{10^{-1}}\right)^2}\)

⇒ \(\frac{\mu_0}{10^{-1}} \sqrt{9+16}=\frac{5 \mu_0}{10^{-1}}\)

= 50 x 4π x 10^-7

= 62.8 x 10^-6 T

= 62.8μT

The direction of the resulting magnetic field,

⇒ \(\tan \theta=\frac{B_2}{B_1}=\frac{3 \mu_0 \times 10^{-1}}{4 \mu_0 \times 10^{-1}}\)

or, \(\tan \theta=\frac{3}{4} \quad \text { or, } \theta \approx 37^{\circ}\)

Resultant magnetic field B makes an angle of 37° with B₁