Test: MCQs (One or More Correct Option): Moving Charges & Magnetism | JEE Advanced - JEE MCQ

The force on north pole = 

The force on south pole =

Since the forces will be unequal and are not having same line of action therefore, the magnetic needle experiences a force as well as a torque.

There is no change in velocity. It can be possible when 

Electric and magnetic fields are absent, i.e., E = 0, B = 0 

Or when electric and magnetic fields are present but force due to electric field is equal and opposite to the magnetic force, (i .e., E ≠ 0, B ≠ 0). 

Or when E = 0. B ≠ 0 provided

F = qvB sin θ = 0

sin θ = 0, i.e., θ = 0 ⇒ v and B are in the same direction.

AB part of the rectangular loop will get attracted to the long straight wire as the currents are parallel and in the same direction whereas CD part will be repelled. But since this force where r is the distance between the wires. Therefore, there will be a net attractive force on the rectangular loop. Force on BC is equal and opposite to that on AD.

The magnetic field due to current in wire 1 in the region of wire 2 will be

Since wire 2 having current i is placed in a magnetic field B1, it will experience a force given by F = i (lB₁ sin 90°)

K.E. of first particle

K.E. of second particle

NOTE : After entering the magnetic field, a magnetic force acts on the charged particle which moves the charged particle in circular path of radius

Here, K, q, B are equal

Considering the activity from P to Q (Horizontal) u₁ = v, v₁ = 2v, s₁ = 2a, Acc = A

⇒ 4v² – v² = 2A(2a)

Force acting in the horizontal direction is F = qE = mA

Rate of doing work at P

Rate of doing work by the magnetic field is throughout zero.The rate of doing work by electric field is zero at Q. Because at Q, the angle between force due to electric field and displacement is zero.

For V = Ig (G + R) = 5 × 10^–5 [100 + 200,000] = 10V

Let us consider any poin t P inside the thin walled pipe. Let us consider a circular loop and apply Ampere's circuital law,

Since current inside the loop is zero.

KEY CONCEPT : When the charged particles enter a magnetic field then a force acts on the particle which will act as a centripetal force. We know that when kinetic energy and magnetic field are equal then

He⁺ and O⁺⁺ will be deflected equally.
H⁺ will be deflected the most since its radius is smallest.

Current, i = (frequency) (charge)=

Magnetic moment,

Angular momentum .

Net magnetic field due to the wires will be downward as shown below in the figure. Since angle between and  is 180°,

Therefore, magnetic force 

Out of the given options only induced electric field and magnetostatic field form closed loops of field lines.

Net force on the loop : Force on AB : The magnetic field due to current I₁ is along AB.

dF = I (d l * B ´ sin 0°)=0

Force on CD : Similarly the magnetic field due to current I₁ is along DC. Because θ = 180° here, therefore force on DC is zero.

Force on BC : Consider a small element dl.

By Fleming's left hand rule, the direction of this force is perpendicular to the plane of the paper directed outwards.

By Fleming's left hand rule, the direction of this force is perpendicular to the plane of paper directed inwards. Since the current elements are located symmetrical to current I₁, therefore force on BC will cancel out the effect of force on AD.
⇒ Net force on loop ABCD is zero.
Net Torque on the loop : The force on BC and AD will create a torque on ABCD in clockwise direction about OO' as seen by the observer at O.

As the particle enters the magnetic field, a force acts on it due to the magnetic field which moves the particle in a circular path of radius

For the particle to enter region III, r > l (path shown by dotted line)

For maximum path length in region  II, r = l

The time taken by the particle to move in region II before coming back in region I is given by

When current I is switched on in both the solenoids in identical manner, eddy currents are setup in metallic rings A and B in such a way that rings A and B are repelled.

Given h_A >h_B . This shows that eddy currents produced in A are greater than in B. This is possible when r A <rB (the rate of change of flux is same in both the rings, therefore induced emf is same).

Figure shows that the megnetic field is present on the right hand side of AB. The electron (e) and proton (p) moving on straight parallel paths with the same velocity enter the region of uniform magnetic field.
The entry and exit of electron & proton in the magnetic field makes the same angle with AB as shown.
Therefore both will come out travelling in parallel paths.

The time taken by proton

The time taken by electron is

∴ (b) , (d)  are correct options

When θ = 0º, the charged particle is projected along xaxis, due to the charged particle will tend to move in a circular path in y-z plane but due to force of electric field, the particle will move in a helical path with increasing pitch. Therefore options (A) and (B) are incorrect.

When θ = 10º, we can resolve velocity into two rectangular components. One along x-axis (v cos 10º) and one along yaxis (v sin 10º). Due to v cos 10º, the particle will move in circular path and due to v sin 10º plus the force due to electric field, the particle will undergo helical motion with its pitch increasing.
If θ = 90º, the charge is moving along the magnetic field.
Therefore the force due to magnetic field is zero. But the force due to electric field will accelerate the particle along yaxis.

The magnetic field should be in the –z direction (Fleming’s left hand rule)

(a) and (c) are the correct options

In the region O < r < R, the magnetic field is present due to current in solenoid.

In the region r > 2 R, the magnetic field is present due to the current in the cylinder.
For the region R < r < 2R, the magnetic field is neither along the common axis, nor tangential to the circle of radius r. (a) and (d) are correct options.

option [A] is correct

Option (b) and (c) are also correct

The range of voltmeter ‘V’ is given by the expression

V is max in this case as RHS is maximum. Thus (a) is correct.
The range of ammeter I is given by the expression

Here R_eq is minimum and therefore I is maximum. Thus (c) is the correct option.