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This contains 10 Multiple Choice Questions for Physics Magnetostatics MCQ Level - 2 (mcq) to study with solutions a complete question bank.
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QUESTION: 1

Figure here shows three close in all cases the circular path has radius ** r** and straight ones are infinitely long. For same current the magnetic field at the centre

Solution:

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)

Magnetic field at

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:

QUESTION: 2

A disc of radius ** R** rotates constant angular velocity ω about its own axis surface density of this disc varies as σ = αr

Solution:

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

QUESTION: 3

For a solenoid of length ** l**, having

Solution:

Let a solenoid of length ** l**, having

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

QUESTION: 4

A particle of mass ** m** and charge

Solution:

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

which is independent of

The correct answer is: The particle enter Region III only if its velocity

QUESTION: 5

The magnetic field at the centre of square of side * a* is :

Solution:

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:

QUESTION: 6

A coil of 100 turns and area **2 × 10 ^{–2} m^{2}** is pivoted about a vertical diameter in a uniform magnetic field and carries a current of 5

Solution:

Torque due to magnetic field,

** N, B, i** and

From equation (i)

The correct answer is: 0.05 *T*

QUESTION: 7

A wire of length * L *is bent in the form of a circular coil and current

Solution:

The are ** n** turns having

The correct answer is: 1

QUESTION: 8

Which of the following statement is (are) correct in the given figure?

Solution:

We know that

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

In case of wire

because magnetic lines are anti parallel to this wire.

In case of ** CB**, and

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

The correct answer is: loop will rotate clockwise about axis

QUESTION: 9

A charge q is moving with a velocity at a point in a magnetic field and experience a force If the charge is moving with velocity at the same point, it experience a force

F_{2} = q(i - k)N The magnetic induction * B* at that point is :

Solution:

Let the magnetic field is,

Applying ** F_{m} = q(V × B)** two times, we have

Comparing two sides, we get

Again comparing we get,

QUESTION: 10

Two circular coil 1 and 2 are made from the same wire but the radius of the 1st coil is twice that of the 2nd coil. What is the ratio of potential difference applied across them, so that magnetic field at their centre is same ?

Solution:

Magnetic field due to current in coil-1

Magnetic field due to current in coil-2

But *B*_{1} = *B*_{2}

Now, ratio of potential difference

The correct answer is: 4

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