JEE Advanced (Single Correct MCQs): Moving Charges & Magnetism - JEE MCQ

The magnetic field is perpendicular to the plane of the paper. Let us consider two diametrically opposite elements. By Fleming's left hand rule, on element AB the direction of force will be leftwards and the magnitude will be 

dF = I ( d l)Bsin 90°= I ( dl)B

On element CD, the direction of force will be towards right on the plane of the paper and the magnitude will be

dF = I ( d l)B.
These two forces will cancel out.
NOTE : Similarly, all forces acting on the diametrically opposite elements will cancel out in pair. The net force acting on the loop will be zero.

Magnetic field at the centre due to current in arc ABC is

Magnetic field at the centre due to current in arc ADB is

Therefore net magnetic field at the centre

KEY CONCEPT :

The magnetic lines of force created due to current will be in such a way that on x – y plane these lines will be perpendicular. Further, these lines will be in circular loops. The number of lines moving downwards in x – y plane will be same in number to that coming upwards of the x – y plane. Therefore, the net flux will be zero.
One such magnetic line is shown in the figure.

Force due to electric field will make the charged particle released from rest to move in the straight line (that of electric field). Since the force due to magnetic field is zero, therefore, the charged particle will move in a straight line.

The angular momentum L of the particle is given by

The wires at A and B are perpendicular to the plane of paper and current is towards the reader. Let us consider certain points.
Point C (mid point between A and B) : The magnetic field at C due to A ( ) is in upward direction but magnetic field at C due to B is in downward direction.
Net field is zero.
Point E : Magnetic field due to A is upward and magnetic field due to B is downward but 

∴Net magnetic field is in downward direction.

Net field upwards. Similarly,, other points can be considered.

Case 1 : Magnetic field at M due to PQ and QR is

Case 2 : When wire QS is joined.
H₂ = (Magnetic field at M due to PQ) + (magnetic field at M due to QR) + (Magnetic field at M due to QS)

NOTE : The magnetic field due to an infinitely long wire carrying current at a distance R from the end point is half that at a distance R from the middle point.

Case of positively charged particle : Two forces are acting on the positively charged particle (a) due to electric field in the positive x-direction. (b) Force due to magnetic field.

This forces will move the positively charged particle towards Y-axis.

Case of negatively charged particle.
Two forces are acting on the negatively charged particle (a) due to electric field in the negative X-direction. (b) due to magnetic field

Same direction as that of positive charge. (c) is the correct answer.

NOTE : If we take individual length for the purpose of calculating the magnetic field in a 3-Dimensional figure then it will be difficult.
Here a smart choice is divide the loop into two loops.
One loop is ADEFA in y-z plane and the other loop will be ABCDA in the x – y plane.

We actually do not have any current in the segment AD. By choosing the loops we find that in one loop we have to take current from A to D and in the other one from D to A. Hence these two cancel out the effect of each other as far as creating magnetic field at the concerned point P is considered.
The point (a, 0, a) is in the X-Z plane.
The magnetic field due to current in ABCDA will be in + ve Z-direction.

NOTE : Due to symmetry the y-components and xcomponents will cancel out each other.
Similarly the magnetic field due to current in ADEFA will be  in x-direction.

∴ The resultant magnetic field will be

KEY CONCEPT : When a charged particle is moving at right angles to the magnetic field then a force acts on it which behaves as a centripetal force and moves the particle in circular motion.

Let us consider a thickness dx of wire. Let it be at a distance x from the centre O.

Number of turns per unit length

∴ Number of turns in thickness dx =dx

Small amount of magnetic field is produced at O due to thickness dx of the wire.

On integrating, we get,

Width of the magnetic field region where 'R' is its radius of curvature inside magnetic field,

The wire carries a current I in the negative z-direction.
We have to consider the magnetic vector field

Magnetic field is perpendicular to OP.

NOTE : Magnetic lines of force form closed loops.
Inside a magnet, these are directed from south to north pole.

The velocity at P is in the X-direction (given).

After P, the positively charged particle gets deflected in the x – y plane toward – y direction and the path is non-circular.

Since in option (b), electric field is also present therefore it will also exert a force in the + X direction.
The net result of the two forces will be a non-circular path.
Only option (b) fits for the above logic. For other option, we get some other results.

KEY CONCEPT : Use Fleming's left hand rule. We find that a force is acting in the radially outward direction throughout the circumference of the conducting loop.

As we know that, potential energy of a magnet in a magnetic field
U=−m⋅B
  =−mBcosθ
where, m= magnetic dipole moment of the magnet 
B= magnetic field

The force acting on electron will be perpendicular to the direction of velocity till the electron remains in the magnetic field. So the electron will follow the path as given.

Use the vector for m of B and v in the formulae to get the instantaneous direction of force at x = a and x = 2a.

Let us consider an elemental length dl subtending an angle dθ at the centre of the circle. Let F_B be the magnetic force acting on this length. Then

F_B = BI (dl)  directed upwards as shown

= BI (Rdθ)              

Let T be the tension in the wire acting along both ends of the elemental length as shown. On resolving T, we find that the components. cancel out and
the components.  add up to balance FB.
At equilibrium  

The magnetic moment of a current carrying loop is given by

the direction is towards positive z-axis.