The document Previous year questions (2016-20) - Moving Charges and Magnetism (Part - 1) Notes | EduRev is a part of the JEE Course Class 12 Physics 35 Years JEE Mains &Advance Past year Paper.

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**Q 1. The figure gives experimentally measured B vs H variation in a ferromagnetic material. The retentivity, coercivity and saturation, respectively, of the material are [2020](1) 1.5 T, 50 A/m and 1.0 T (2) 1.5 T, 50 A/m and 1.0 T (3) 150 A/m, 1.0 T and 1.5 T (4) 1.0 T, 50 A/m and 1.5 T Ans:** (4)

From the given graph, we have

Retentivity = 1.0 T

It is the ability of a substance to retain or resist magnetization.

Coercivity = 50 A/m

It is the resistance of a magnetic material to changes in magnetization.

And Saturation = 1.5 T

Given, circular coil curving current and forming magnetic dipole.

ϕ

ϕ

Since, magnetic field forms closed loop so magnetic flux passing through area of circular coil and outside must be equal but opposite in direction

Therefore,

ϕ_{i} = -ϕ_{0}**Q 3. A loop ABCDEFA of straight edges has six corner points A(0, 0, 0), B(5, 0, 0), C(5, 5, 0), D(0, 5, 0), E(0,5, 5) and F(0, 0, 5). The magnetic field in this region is The quantity of flux through the loop ABCDEFA (in Wb) is ______. [2020]Ans: **175

Given that

Corner points of loop ABCDEFA are A(0, 0, 0), B(5, 0, 0), C(5, 5, 0), D(0, 5, 0), E(0, 5, 5) and F(0, 0, 5).

Area vector of loop is

Magnetic flux is given by

Given that

Force acting on charge along x-axis is

So, acceleration is

(1)

We need to find the time when speed will be double, that is,

From Eq. (1), we get

(1) 0.71 mT

(2) 7.1 mT

(3) 0.071 mT

(4) 71 mT

Ans:

Given that

K = 1MeV = 1 x 10

m

a = 10

Force on a moving charge particle in a magnetic field is given by

F qvB = sinθ

Here θ = 90^{0 }

So,

F = qvB

and F = ma

So, ma = qvB

We know that

Then,**Q 6. A very long wire ABDMNDC is shown in figure carrying current I. AB and BC parts are straight, long and at a right angle. At D, wire forms a circular turn DMND of radius R. AB, BC parts are tangential to circular turn at N and D. Magnetic field at the centre of circle is [2020]****Ans: **(1)**Solution:**

There are three magnetic fields act at the centre of the circle, that is

Magnetic field due to straight part of wire AB, that is,

Since, perpendicular distance between wire part AB and centre of circle is R. Direction of field of inside the paper.

Magnetic field due to circular part of wire DMND, that is,

Direction of field is outside the paper.

Magnetic field due to straight part of wire BDC, that is,

Since, perpendicular distance between wire part BDC and centre of circle is R. Direction of field is outside the paper.

Therefore, total magnetic field act at the centre of the circle DMND is given by

Since, we are taking the direction outside the paper as reference so the inside magnetic field become negative.

So,**Q 7. A galvanometer having a coil resistance 100 Ω gives a full scale deflection when a current of 1 mA is sped through it. What is the value of the resistance which can convert this galvanometer into a voltmeter giving full scale deflection for a potential difference of 10 V? [2020](1) 10 kΩ(2) 8.9 kΩ(3) 7.9 kΩ(4) 9.9 kΩAns:** (4)

Given that

R

To make galvanometer to voltmeter, we add a resistance in series

So,

V = I

(1) 2/3

(2) 2

(3) 1/2

(4) 3/2

Ans:

Given condition is shown in the following circuit figure

Let current density of the given wire b J

Applying Ampere's law, we have

At point A distance a/3, we get

(1)

At point B distance 2a , we get

(2)

From Eqs. (1) and (2), we get

**Q 9. A charged particle of mass m and charge q moving under the influence of uniform electric field and a uniform magnetic field follows a trajectory from point P to Q as shown in figure. The velocities at P and Q are respectively, Then which of the following statements (A, B, C, D) are correct? (Trajectory shown is schematic and is not to scale) [2020](B) Rate of work done by the electric field at P is (C) Rate of work done by both the fields at Q is zero(D) The difference between the magnitude of angular momentum of the particle at P and Q is 2 mav.(1) (A), (C), (D) (2) (B), (C), (D) (3) (A), (B), (C) (4) (A), (B), (C), (D)**

Solving for point (A):

From work energy theorem, we have

W

So, point (A) is true.

Solving for point (B):

Rate of work done by electric filed = Power of electric field

Power of electric filed = F

So, point (B) is true.

Solving for point (C):

At point Q,

Total rate of work done = Rate of work done by electric filed + Rate of work done by magnetic field

Rate of work done by magnetic field = 0 W (Since, force and displacement are perpendicular)

Rate of work done by electric field = 0 W (Since, force and displacement are perpendicular)

Therefore, total rate of work done at point Q by both fields = 0 W

So, point (C) is true

Solving for point (D):

Initial angular momentum is

L

Final angular momentum is

L

Change in angular momentum is

ΔL = L

So, point (D) is not true.

**Ans: **(2)**Solution:**

The maximum possible radius of electron for which it does not hit the surface is

r = R/2 .. (1)

Radius of circular motion of a charge particle in a magnetic field is given by

r = mv/qB .. (2)

From Eqs. (1) and (2), we get

Since, magnetic field due to solenoid is given by

B = μ_{0}nI

**Q 11. A current loop, having two circular arcs joined by two radial lines is shown in the figure. It carries a current of 10 A. The magnetic field at point O will be close to [2019]**

**(1) 1.0 × 10 ^{−7} T**

Ans:

Let B be the magnetic field at point O due to are of radius r.

Given r

Therefore,

Since, both arcs have opposite direction of current

Ans:

Let the effective length of magnet be l.

So, magnetic moment is

M = i

= i

Suppose i_{w} is the current flowing through the wire. The force on north pole of magnet = mB and the force on south pole of magnet = mB. Then, resultant force is

F = 2mBcosθ

l << d, so l can be neglected.

Therefore,**Q 13. 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, the mass of the particle is (Given charge of electron = 1.6 × 10 ^{−19} C) [2019](1) 9.1 × 10^{−31} kg (2) 16 × 10^{−27} kg (3) 1.6 × 10^{−19} kg (4) 2.0 × 10^{−24} kg Ans:** (4)

The force acting on the charge particle due to magnetic field is

The force F on the charged particle due to magnetic field provides the required centripetal force necessary for motion along the circular path of radius r

From Eq. (1) and Eq. (2), we get

Force due to electric field is

F = qE

Since, both are in straight line, then

qvB = qE

⇒ E = vB (4)

From Eq. (3) and Eq. (4), we have

(1) N

(2) 1/N

(3) N

Ans:

Let R be the radius of loop and r be the radius of circular coil of N turn.

For Loop,

L = 2πR (1)

For coil,

L = N × 2πr (2)

Fron Eq.(1) and Eq. (2), we get

R = Nr

Magnetic field at the centre of loop is

Magnetic field at the centre of coil is**Q 15. 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) [2019]**(4)

(1) 7.5 × 10^{−3} m

(2) 7.5 m

(3) 7.5 × 10^{−2} m

(4) 7.5 × 10^{−4} m

Ans:

We know

(1) 100 Ω

(2) 120 Ω

(3) 80 Ω

(4) 125 Ω

Ans:

Deflection current is

i

= 30 × 0.005

Thus, V = IR

⇒ 15 = (0.05 × 30 × 20) + (30 × 0.005 × R)

⇒ 15 = 3 + 0.15R**Q 17. The region between y = 0 and y = d contains a magnetic field A particle of mass m and charge q enters the region with a velocity the acceleration of the charged particle at the point of its emergence at the other side is [2019]Ans:** (*)

*Disputed question – None of the options is correct.

Let us suppose that particles enter from circular path with center (0, d).

Radius of circular path is**Q 18. A particle of mass m and charge q is in an electric and magnetic field given by The charged particle is shifted from the origin to the point P (x = 1; y = 1) along a straight path. The magnitude of the total work done is [2019](1) (0.35)q (2) (5)q (3) (2.5)q (4) (0.15)q**

Electric and magnetic field is given by

Work done by magnetic force = 0

So,

Therefore, work done by electric field

**(1) 5 Ω(2) 22 Ω(3) 25 Ω(4) 12 ΩAns:** (2)

Let R

Case II

(2)

⇒ 225R

⇒ 225R

⇒ 200R

(1) 250 Ω

(2) 200 Ω

(3) 6200 Ω

(4) 6250 Ω

Ans:

= 10

Now,

2.5 = (50 + R) × I

⇒ 2.5 = (50 + R) × 10

⇒ 2.5 × 10

⇒ 250 = 50 + R

⇒ R = 200 Ω

(1) 285 A/m

(2) 2600 A/m

(3) 520 A/m

(4) 1200 A/m

Ans:

Bar magnet requires a magnetic intensity μ.

The current through the solenoid is,

i = μ/n.

⇒ μ = i x n (1)

Let n be the turns per unit length.

Put this value in Eq. (1), we get

(1) 0.01 J

(2) 0.007 J

(3) 0.028 J

(4) 0.014 J

Ans:

*Disputed question – None of the options is correct.

B = cos(0.125)

Therefore, Work done is

(1) πnρl

(4) nρl

Magnetic moment M = NIA

Charge of element dq = ρdx

Magnetic moment of this charge element =

Integrating both the sides, we get

(1) T

(2) T

(3) T

(4) T

Ans:

Let μ be the magnetic moment, M masses of hoop and solid cylinder and R is the radius of hoop and solid cylinder.

Oscillation period of hoop is

(1) [T_{h }= MR^{2}]

Oscillation period of cylinder is

Dividing Eq. (1) by Eq. (2), we get**Q 25. At some location on Earth, the horizontal component of Earth’s magnetic field is 18 × 10 ^{−6} T. At this location, magnetic needle of length 0.12 m and pole strength 1.8 A-m is suspended from its mid-point using a thread, it makes 45° angle with horizontal in equilibrium. To keep this needle horizontal, the vertical force that should be applied at one of its ends is [2019](1) 3.6 × 10^{−5} N(2) 1.8 × 10^{−5} N(3) 1.3 × 10^{−5} N(4) 6.5 × 10^{−5} NAns:** (4)

We have,

.

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