Magnetostatics MCQ Level - 2 - IIT JAM MCQ

For first figure magnetic field at O  due to straight part (a)

Magnetic field at O  due to circular part (b)

Magnetic field at O  due to straight part (c)

So, net magnetic field at the centre of first figure

For second figure

Magnetic field at O  due to two straight and one circular path is

For third figure magnetic field at O  due to straight part (a)
B_a = 0
Magnetic field at O  due to circular part (b)

Magnetic field at C due to straight part  (c)

So net magnetic field at centre of third figure

From equation (i), (ii) and (iii)

The correct answer is: 

To find the magnetic field due to charge on rotating disc. Consider a ring of radius r and thickness dr.

So, dA = 2πrdr

Charge on element 

Magnetic field at the centre due to charge element

Let a solenoid of length l, having N  turns, is given as figure. To find magnetic field at a point on its axis, consider an element of a solenoid in the form of ring of radius a (same as solenoid) and thickness dx.

Let n  be the number of turns per unit length.

Number of turns in dx length = dN = ndx
Let I  be the current flowing in the soil.

Using result of the ring, the magnetic field at the point P  is given by

In region II,

Magnetic field  is perpendicular to velocity  therefore, path of particle is circle is region II Particle enters in region III if, radius of circular path,

r > l

 particle will turn back and path length will be maximum. If particle returns to region I, time spent in region II will be :
 which is independent of V
The correct answer is: The particle enter Region III only if its velocity 

Magnetic field due to a current carrying conductor at a point on the perpendicular axis passing through mid point is given by
   α and β are the angle

So,

The correct answer is: 

Torque due to magnetic field,

N, B, i  and  A  are constant

From equation (i)

The correct answer is: 0.05 T

The are n  turns having i  current each, so  I = ni


n_min = 1
The correct answer is: 1

We know that

In case of wire BA :  because magnetic line are parallel to this wire.
In case of wire CD,  

 because magnetic lines are anti parallel to this wire.

In case of  CB, and  AD,
 is perpendicular to paper outward and  is perpendicular to paper inward. These two force (although calculated by integration) cancel each other but produce a torque which tend to relate the loop in clockwise direction about an axis OO'.
The correct answer is: loop will rotate clockwise about axis OO'  when seen from O

Let the magnetic field is,

Applying  F_m = q(V × B)  two times, we have

Comparing two sides, we get B₂ = 1 and B₃ = 1

Again comparing we get, B₁ = 1 and B₃ = 1

Magnetic field due to current in coil-1

Magnetic field due to current in coil-2

But  B₁ = B₂

Now, ratio of potential difference

The correct answer is: 4