Question 1

A tightly wound solenoid has 2000 turns and a cross-sectional area of 1.5 × 10⁻⁴ m². It carries a current of 2.0 A and is hung at its center with its axis perpendicular to its length so that it can rotate in a horizontal plane. If it is placed in a uniform magnetic field of strength 5 × 10

Accepted Answer

1.5 × 10⁻² N-m. tau = N*I*A*B*sin(30) = 2000 * 2.0 * 1.5e-4 * 5e-2 * 0.5 = 1.5e-2 N-m.

Question 2

A cyclotron with dee radius R is used to speed up protons of mass m. An alternating electric field of frequency ν is applied across the dees. For the cyclotron to operate, the magnetic field B and the kinetic energy K of the emerging proton beam are related by:

Accepted Answer

B = 2πmν / e and K = 2mπ²ν²R². Cyclotron resonance: nu = eB/(2 pi m) so B = 2 pi m nu/e. v_max = 2 pi nu R, so K = (1/2)m(2 pi nu R)^2 = 2 m pi^2 nu^2 R^2.

Question 3

A long, straight conductor of radius a carries a constant current I that is spread uniformly over its cross-sectional area. What is the ratio of the magnetic field at a distance a/2 from the axis to the magnetic field at a distance 2a from the axis?

Accepted Answer

1. The magnetic field inside a long, straight conductor increases linearly with distance from the center, while outside the conductor, it decreases with the distance from the axis. At a distance a/2, the magnetic field is proportional to the distance from the center, while at 2a, it behaves inversely with distance, leading to a ratio of 1.

Question 4

A charge of 2 μC revolves with a frequency of 6.25 × 10¹² Hz and creates a magnetic field of 6.28 T at a point on the circle. What is the radius of the circular path?

Accepted Answer

1.25 m. I = q*f = 2e-6 * 6.25e12 = 1.25e7 A. From B = mu0*I/(2r), r = mu0*I/(2B) = (4*pi*1e-7 * 1.25e7)/(2*6.28) = 1.25 m.

Question 5

An electron beam travels through a region where uniform electric and magnetic fields act mutually perpendicular to each other, and both are also perpendicular to the direction of motion of the electrons. If the electric field strength is 20 V m−1 and the magnetic field strength is 0.5 T, t

Accepted Answer

40 m/s. For undeflected motion the electric and magnetic forces balance: qE = qvB, so v = E/B = 20/0.5 = 40 m/s.

Question 6

A particle having mass m and charge q is first accelerated through a potential difference V and then enters a zone of uniform magnetic field B acting perpendicular to its velocity. If the magnetic field extends over a width d, what is the angle through which the particle’s direction change

Accepted Answer

sin−1[ Bd (q/2mV)1/2 ]. From qV=(1/2)mv^2, v=sqrt(2qV/m); radius r=mv/(qB). Deflection angle theta has sin(theta)=d/r = Bd*(q/(2mV))^(1/2), so theta = sin^-1[Bd(q/2mV)^(1/2)].

Question 7

A charged particle of charge q and mass m is moving in a circular path of radius r with angular velocity ω. The ratio of the magnitude of its magnetic moment to the magnitude of its angular momentum depends on

Accepted Answer

q and m. Magnetic moment = (1/2)*q*omega*r^2 and angular momentum = m*omega*r^2, so their ratio = q/(2m), which depends only on q and m.

Question 8

In a mass spectrometer, ions are first accelerated through a potential difference V and then forced by a magnetic field B to move along a semicircular path of radius R. If V and B remain unchanged, how does the charge-to-mass ratio of the ion vary with R?

Accepted Answer

It is proportional to 1/R². From qV = (1/2)mv^2 and R = mv/(qB): q/m = 2V/(R^2 B^2), so q/m is proportional to 1/R^2.

Question 9

A long straight wire carrying a constant current is shaped first into a single-turn circular loop, producing a magnetic field of magnitude B at its centre. If the same wire, with the same current, is instead wound into a circular coil of n turns having a smaller radius, what will be the ma

Accepted Answer

n²B. When the wire is wound into a coil with n turns, the magnetic field at the center increases because each turn contributes to the total magnetic field. The magnetic field strength is proportional to the number of turns squared, resulting in a field strength of n²B at the center.

Question 10

A charged particle enters a magnetic field with its velocity at right angles to the field. Its motion will be such that

Accepted Answer

its momentum varies, while its kinetic energy remains unchanged. The magnetic force is perpendicular to velocity, so it does no work: kinetic energy (and speed) stay constant. Momentum is a vector whose direction continuously changes, so momentum varies.

Question 11

A charged particle starts from rest in a region where uniform, time-independent electric and magnetic fields are present and both fields point in the same direction. The particle’s path will be

Accepted Answer

a straight path. From rest, only E acts initially, accelerating the particle along B. Since the velocity stays parallel to B, the magnetic force qv x B is always zero, so the path is a straight line along the field.

Question 12

An equilateral triangular coil of side l is placed between the pole pieces of a permanent magnet so that the magnetic field B lies in the plane of the coil. If a current i flowing through the triangle produces a torque τ on it, then the side length l of the triangle is

Accepted Answer

2(τ/√3Bl)^(1/2). With B in the coil's plane, torque tau = i*A*B = i*(sqrt3/4)*l^2*B. Solving, l = 2*sqrt(tau/(sqrt3 * B * i)), which is index 1.

Question 13

A magnetic needle is placed in a magnetic field that is not uniform. What will it experience?

Accepted Answer

Both force and torque. In a non-uniform magnetic field, the magnetic needle experiences a gradient in the magnetic field strength, which results in both a force acting on it and a torque that tends to align it with the field. The force arises from the variation in magnetic field strength across the needle, while the torque is due to the interaction of the needle's magnetic moment with the field.

Question 14

A dead-beat galvanometer shows a pointer that quickly comes to rest at a fixed reading because

Accepted Answer

eddy currents are set up in the metallic frame supporting the coil.. A dead-beat galvanometer has its coil wound on a metallic (conducting) frame; the induced eddy currents oppose the motion, quickly damping oscillations so the pointer settles fast.

Question 15

When an electric current flows through a spring, what happens to it?

Accepted Answer

It contracts. When an electric current flows through a spring, it generates heat due to resistance, causing the material to expand initially but then contract as it cools down, resulting in an overall contraction.

Question 16

A circular loop A has radius R and carries a current I. Another circular loop B has radius 2R and carries a current 2I. What is the ratio of the magnetic field strengths produced at their centers, B_A: B_B?

Accepted Answer

1. The magnetic field at the center of a circular loop is directly proportional to the current and inversely proportional to the radius. For loop A, the field is proportional to I/R, and for loop B, it is proportional to (2I)/(2R). Simplifying both expressions shows that they are equal, resulting in a ratio of 1.

Question 17

An electron and a proton, each with the same momentum, enter a uniform magnetic field at right angles to the field. What will be the nature of their paths?

Accepted Answer

Their trajectories will have the same curvature, apart from the direction of rotation.. Both the electron and the proton experience the same magnetic force due to their equal momentum and perpendicular entry into the magnetic field, resulting in circular paths. However, the electron, being lighter, will have a faster angular velocity, leading to the same curvature but opposite rotation direction.

Question 18

For a charged particle moving in a circle inside a uniform magnetic field, the time period does not depend on its

Accepted Answer

speed. The time period of a charged particle moving in a magnetic field is determined by the radius of the circular path and the magnetic field strength, and it remains constant regardless of the particle's speed. This is because the magnetic force provides the necessary centripetal force, leading to a fixed frequency of rotation.

Question 19

A particle having mass M and charge Q moves with speed v in a uniform magnetic field B applied perpendicular to its velocity, so that it traces a circle of radius R. The total work done by the magnetic field during one complete revolution is

Accepted Answer

zero. The work done by a magnetic field on a charged particle is always zero because the magnetic force acts perpendicular to the velocity of the particle, resulting in no displacement in the direction of the force.

Question 20

A long conductor carries a constant current. When it is shaped into a single-turn circular loop, the magnetic field at the centre is B. If the same wire is then formed into a circular coil having n turns, the magnetic field at the centre becomes

Accepted Answer

n²B. Fixed wire length L = 2*pi*R. With n turns each radius r = R/n. B' = n*mu0*I/(2r) = n*mu0*I/(2R/n) = n^2 * mu0*I/(2R) = n^2 B.

Question 21

A circular wire loop of radius 3 cm carries a current. The magnetic field produced by it at a point on its axis 4 cm from the centre is 54 μT. What is the magnetic field at the centre of the loop?

Accepted Answer

250 μT. B_centre = B_axis*(R^2+x^2)^(3/2)/R^3 = 54*(3^2+4^2)^(3/2)/3^3 = 54*125/27 = 250 microT.

Question 22

Two very long parallel wires are a distance d apart and carry currents I1 and I2 in the same direction, producing a mutual force of magnitude F. If the current in one wire is changed to twice its original value and its direction is reversed, while the separation is increased to 3d, what is

Accepted Answer

-2F/3. F = mu0*I1*I2/(2*pi*d). Doubling one current and tripling the separation scales the magnitude by 2/3, and reversing that current flips attraction to repulsion (sign change). New force = -2F/3.

Question 23

Two circular coils are concentric, each having a radius of 2π cm, and the planes of the coils are mutually perpendicular. If currents of 3 A and 4 A flow through the two coils, respectively, what is the magnitude of the magnetic induction at the common centre? Take μ0 = 4π × 10⁻⁷ Wb/A·m.

Accepted Answer

5 × 10⁻⁵. The magnetic induction at the center of two perpendicular coils can be calculated using the formula for the magnetic field due to a circular coil, which is proportional to the current and inversely proportional to the radius. By applying the formula for both coils and considering their perpendicular arrangement, the resultant magnetic induction is found to be 5 × 10⁻⁵ Wb/m².

Question 24

A particle with mass m and charge q moves in a circle of radius r in a plane perpendicular to a magnetic field of magnitude B. The period of one complete revolution is

Accepted Answer

2πm/qB. The period of revolution for a charged particle in a magnetic field is derived from the balance of the magnetic force and the centripetal force, leading to the formula T = 2πm/(qB), where T is the period, m is the mass, q is the charge, and B is the magnetic field strength.

Question 25

Two very long, thin, parallel conductors are placed a distance d apart. If each carries the same current i in the same direction, then they will

Accepted Answer

attract one another with a force of μ0i²/(2πd). Two long parallel wires carrying current in the same direction attract each other, with force per unit length F/L = mu0*i^2/(2*pi*d).

Question 26

A charged particle is set free from rest in a region where constant, uniform electric and magnetic fields exist and both fields point in the same direction. The particle will move along a

Accepted Answer

straight path. The charge starts at rest, so qE accelerates it along E, which is parallel to B. Its velocity stays parallel to B, making qv x B = 0 at all times. It therefore moves in a straight line.

Question 27

A long solenoid has a winding density of 200 turns per cm and carries a current i. The magnetic field at its centre is 6.28 × 10⁻² Weber/m². A second long solenoid has 100 turns per cm and carries a current i/3. What magnetic field is produced at its centre?

Accepted Answer

1.05 × 10⁻² Weber/m². B = mu0*n*i, so B2 = B1*(n2/n1)*(i2/i1) = 6.28e-2*(100/200)*(1/3) = 1.05e-2 Wb/m^2.

Question 28

A long, straight conductor of radius a carries a constant current i that is spread uniformly over its cross-sectional area. What is the ratio of the magnetic field magnitude at a distance a/2 from the axis to that at a distance 2a from the axis?

Accepted Answer

1. At a/2 (inside): B = mu0*i*(a/2)/(2*pi*a^2) = mu0*i/(4*pi*a). At 2a (outside): B = mu0*i/(2*pi*2a) = mu0*i/(4*pi*a). The two are equal, so the ratio is 1.

Question 29

A current I flows along the length of an infinitely long, straight, thin walled pipe. Then

Accepted Answer

the magnetic field at any point inside the pipe is zero. According to Ampère's law and the properties of magnetic fields generated by current-carrying conductors, the magnetic field inside a hollow conductor is zero due to the symmetry of the current distribution, which results in the cancellation of the magnetic field lines.

Question 30

A charged particle with charge q enters a region of constant, uniform, and mutually orthogonal fields E and B with a velocity v perpendicular to both E and B and comes out without any change in magnitude or direction of v. Then

Accepted Answer

v = E × B / B². The correct option is derived from the condition that the net force acting on the charged particle must be zero for it to maintain its velocity. In a uniform electric field E and magnetic field B, the electric force qE and the magnetic force q(v × B) must balance each other, leading to the relationship v = E × B / B².

Question 31

A charged particle moves through a magnetic field perpendicular to its direction. Then

Accepted Answer

the momentum changes but the kinetic energy is constant. When a charged particle moves through a magnetic field perpendicular to its velocity, the magnetic force acts as a centripetal force, changing the direction of the particle's momentum without altering its speed. Since the speed remains constant, the kinetic energy, which depends on speed, also remains constant.

Question 32

A current I flows in an infinitely long wire with cross section in the form of a semi-circular ring of radius R. The magnitude of the magnetic induction along its axis is:

Accepted Answer

(d) μ0I / π2R. The magnetic induction along the axis of a semi-circular wire can be derived using the Biot-Savart law, which shows that the contribution to the magnetic field from each segment of the wire adds up to yield the result of μ0I / π2R. This reflects the unique geometry of the semi-circular shape, which influences the field strength along the axis.

Question 33

A positive charge +q travels with velocity v = 3î + 4ĵ + k̂ in a region where the electric field is E = 3î + ĵ + 2k̂ and the magnetic field is B = î + ĵ − 3k̂. What is the y-component of the force acting on the charge?

Accepted Answer

11 q. v x B = (3,4,1) x (1,1,-3) = (-13,10,-1). E + v x B = (-10,11,1). The force is q times this, so the y-component is 11q.

Question 34

Proton, deuteron and alpha particle of same kinetic energy are moving in circular trajectories in a constant magnetic field. The radii of proton, deuteron and alpha particle are respectively rₚ, r_d and r_α. Which one of the following relation is correct?

Accepted Answer

(b) r_α = rₚ < r_d. For equal KE, r = sqrt(2mE)/(qB) ~ sqrt(m)/q. Proton: sqrt(1)/1=1; deuteron: sqrt(2)/1=1.41; alpha: sqrt(4)/2=1. Hence r_alpha = r_p < r_d (option b).

Question 35

Two long coaxial solenoids of unequal radii carry the same current I in the same sense. Let F₁ denote the magnetic force on the inner solenoid due to the outer solenoid, and F₂ denote the magnetic force on the outer solenoid due to the inner solenoid. Which statement is correct?

Accepted Answer

F₁ = F₂ = 0. The magnetic fields inside coaxial solenoids do not exert forces on each other when they carry the same current in the same direction, resulting in zero net force on both solenoids.

Question 36

Two wires A and B are identical, each having length l and carrying the same current I. Wire A is shaped into a circular loop of radius R, while wire B is bent into a square of side a. If the magnetic fields at the centers of the circle and the square are BA and BB respectively, what is the

Accepted Answer

π²/(8√2). The magnetic field at the center of a circular loop is given by the formula B = (μ₀I)/(2R), while for a square loop, it can be derived to be B = (μ₀I)/(4a) multiplied by a correction factor due to the geometry. Given that the side length a of the square is related to the radius R of the circle, the ratio BA/BB simplifies to π²/(8√2), making option B the correct answer.

Question 37

An electron, a proton and an alpha particle having the same kinetic energy are moving in circular orbits of radii rₑ, rₚ and r_α respectively in a uniform magnetic field B. The relation between rₑ, rₚ, r_α is

Accepted Answer

rₑ < rₚ = r_α. The radius of the circular orbit in a magnetic field is directly proportional to the momentum of the particle and inversely proportional to its charge. Since the electron has a much smaller mass and charge compared to the proton and alpha particle, it will have a smaller radius for the same kinetic energy, leading to the relation rₑ < rₚ = r_α, as both the proton and alpha particle have the same charge-to-mass ratio.

Question 38

The dipole moment of a circular loop carrying current I, is m and the magnetic field at the centre of the loop is B1. When the dipole moment is doubled by keeping the current constant, the magnetic field at the centre of the loop is B2. The ratio B1/B2 is

Accepted Answer

√2. The magnetic dipole moment of a circular loop is directly proportional to the current and the area of the loop. When the dipole moment is doubled while keeping the current constant, the area must also double, leading to an increase in the magnetic field strength at the center of the loop by a factor of √2, resulting in the ratio B1/B2 being √2.

Question 39

A rectangular coil (Dimension 5 cm × 2.5 cm) with 100 turns, carrying a current of 3 A in the clockwise direction, is kept centered at the origin and in the X-Z plane. A magnetic field of 1 T is applied along X-axis. If the coil is tilted through 45° about Z-axis, the torque on the coil is

Accepted Answer

0.27 Nm. m = N*i*A = 100*3*(0.05*0.025) = 0.375 A m^2. With coil in X-Z plane its normal is along Y (perpendicular to B along X). Tilting 45 deg about Z makes the angle between m and B equal 45 deg, so tau = m*B*sin45 = 0.375*1*0.707 = 0.27 Nm.

Question 40

A current of 1A is flowing on the sides of an equilateral triangle of side 4.5 × 10⁻² m. The magnetic field at the center of the triangle will be -

Accepted Answer

4 × 10⁻⁵ Wb/m². For an equilateral triangle, perpendicular distance from centre to a side d = a/(2*sqrt(3)). Field from one side = (mu0*I/4pi/d)*2*sin60. With a=4.5e-2 m, I=1 A, each side gives ~1.33e-5 T; three sides total ~4e-5 Wb/m^2.

Question 41

A particle having the same charge as of electron moves in a circular path of radius 0.5 cm under the influence of a magnetic field of 0.5 T. If an electric field of 100 V/m makes it to move in a straight path, then the mass of the particle is (Given charge of electron = -1.6 × 10⁻¹⁹ C)

Accepted Answer

2.0 × 10⁻²⁴ kg. Straight-line motion needs qE=qvB so v=E/B=100/0.5=200 m/s. From r=mv/(qB), m=qBr/v=(1.6e-19*0.5*0.005)/200 = 2.0e-24 kg.

Question 42

A proton, an electron, and a helium nucleus have the same energy. They are in circular orbits in a plane due to a magnetic field perpendicular to the plane. Let rₚ, rₑ and r_he be their respective radii. Then:

Accepted Answer

rₑ < rₚ = r_he. For equal energy, r = sqrt(2mE)/(qB) so r is proportional to sqrt(m)/q. For He: sqrt(4m_p)/2 = sqrt(m_p)/1 = same as proton, so r_p = r_he. The electron's tiny mass makes r_e the smallest, giving r_e < r_p = r_he.

Question 43

A square loop is carrying a steady current I and the magnitude of its magnetic dipole moment is m. If this square loop is changed to a circular loop and it carries the same current, the magnitude of the magnetic dipole moment of circular loop will be:

Accepted Answer

4m/π. Square side a gives m = I a^2 with perimeter 4a. Same wire forms a circle: 2*pi*r = 4a => r = 2a/pi, area = pi r^2 = 4a^2/pi. So m_circle = I*(4a^2/pi) = 4m/pi.

Question 44

A rectangular coil (dimension 5 cm × 2.5 cm) with 100 turns, carrying a current of 3 A in the clock-wise direction, is kept centered at the origin and in the X-Z plane. A magnetic field of 1 T is applied along X-axis. If the coil is tilted through 45° about Z-axis, then the torque on the c

Accepted Answer

0.27 N m. The torque on a coil in a magnetic field is calculated using the formula τ = n * I * A * B * sin(θ), where n is the number of turns, I is the current, A is the area of the coil, B is the magnetic field strength, and θ is the angle between the magnetic field and the normal to the coil's plane. In this case, the area of the coil is 0.0125 m², the angle after tilting is 45°, and substituting these values gives the correct torque of 0.27 N m.

Question 45

A moving coil galvanometer has a coil with 175 turns and area 1 cm². It uses a torsion band of torsion constant 10⁻⁶ N-m/rad. The coil is placed in a magnetic field B parallel to its plane. The coil deflects by 10° for a current of 1 mA. The value of B (in Tesla) is approximately:

Accepted Answer

10⁻². The correct option is 10⁻² because the deflection angle of the coil is directly related to the magnetic field strength, the current, and the physical properties of the coil. Using the formula for torque in a galvanometer, we can derive that the magnetic field strength must be approximately 0.01 T to produce a 10° deflection at 1 mA.

Question 46

In an experiment, electrons are accelerated, from rest, by applying a voltage of 500 V. Calculate the radius of the path if a magnetic field 100 mT is then applied. [Charge of the electron = 1.6 × 10⁻¹⁹ C, mass of the electron = 9.1 × 10⁻³¹ kg]

Accepted Answer

7.5 × 10⁻⁴ m. The correct option is right because the radius of the path of an electron in a magnetic field is determined by the balance between the centripetal force and the magnetic force acting on the electron. By calculating the velocity gained from the applied voltage and using the formula for the radius of circular motion in a magnetic field, the result yields a radius of 7.5 × 10⁻⁴ m.

Question 47

A beam of protons with speed 4 × 10⁵ ms⁻¹ enters a uniform magnetic field of 0.3 T at an angle of 60° to the magnetic field. The pitch of the resulting helical path of protons is close to: (Mass of the proton = 1.67 × 10⁻²⁷ kg, charge of the proton = 1.69 × 10⁻¹⁹ C)

Accepted Answer

4 cm. T = 2*pi*(1.67e-27)/(1.69e-19*0.3) = 2.07e-7 s. Pitch = v*cos60*T = 4e5*0.5*2.07e-7 = 0.041 m ~ 4 cm.

Question 48

A charged particle carrying charge 1 μC is moving with velocity (2i + 3j + 4k) m/s. If an external magnetic field of (5i + 3j − 6k) × 10⁻³ T exists in the region where the particle is moving then the force on the particle is F × 10⁻⁹ N. The vector F is-

Accepted Answer

(3) −30i + 32j − 9k. v x B = (-30, 32, -9)*10^-3. F = q(v x B) = 10^-6 * (-30,32,-9)*10^-3 = (-30,32,-9)*10^-9 N, so F = -30i + 32j - 9k.

Question 49

An electron is constrained to move along the y-axis with a speed of 0.1 c (c is the speed of light) in the presence of electromagnetic wave, whose electric field is E = 30 ĵ sin(1.5 × 10⁷ t − 5 × 10⁻² x) V/m. The maximum magnetic force experienced by the electron will be: (given c = 3 × 1

Accepted Answer

4.8 × 10⁻¹⁹ N. B_max = E_max/c = 30/(3e8) = 1e-7 T. v = 0.1c = 3e7 m/s. F = qvB = 1.6e-19 * 3e7 * 1e-7 = 4.8e-19 N.

Question 50

A particle of charge q and mass m is moving with a velocity -v î (v ≠ 0) towards a large screen placed in the Y-Z plane at a distance d. If there is a magnetic field B = B0 k̂, the minimum value of v for which the particle will not hit the screen is:

Accepted Answer

(4) qdB0/m. The particle follows a circular arc whose maximum penetration toward the screen equals its radius r = mv/(qB0). It just fails to reach a screen at distance d when r = d, giving the critical speed v = qdB0/m.

JEE Main Physics: Moving Charges and Magnetism questions with solutions

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