# Magnetic Field Obj-2 HC Verma Solutions Vol-2 Class-12 Ch-34

**Magnetic Field Obj-2 HC Verma Solutions Vol-2 Class-12 Ch-34** Concept of Physics for Class-12. Step by Step Solutions of** Objective -2 (MCQ-2) **Questions of Chapter-34 **Magnetic Field **(Concept of Physics) .Visit official Website CISCE for detail information about ISC Board Class-11 Physics.

## Magnetic Field Obj-2 HC Verma Solutions Vol-2 Class-12 Ch-34

Board | ISC and other board |

Publications | Bharti Bhawan Publishers |

Chapter-34 | Magnetic Field |

Class | 12 |

Vol | 2nd |

writer | HC Verma |

Book Name | Concept of Physics |

Topics | Solution of Objective-2 (MCQ-2) Questions |

Page-Number | 229, 230 |

**-: Select Topics :-**

Question for Short Answer

Obj-1

Obj-2

Exercise

**Magnetic Field Obj-2 (MCQ-2) Questions**

### HC Verma Solutions of Ch-34 Vol-2 Concept of Physics for Class-12

(Page-229)

**Question-1 :-**

If a charged particle at rest experiences no electromagnetic force,

(a) the electric field must be zero

(b) the magnetic field must be zero

(c) the electric field may or may not be zero

(d) the magnetic field may or may not be zero

**Answer-1 :-**

The options **(a) and (d)** are correct

**Explanation:**

As the charged particle is at rest, its velocity,

V = 0 and magnetic force,

F = qVB = 0. Hence,

We cannot determine whether a magnetic field is present or not.

But as the particle at rest experiences no electromagnetic force, the electric field must be zero. This is because electric force acts on a particle whether it is at rest or in motion.

**Magnetic Field Obj-2 (MCQ-2) Questions**

HC Verma Solutions of Ch-34 Vol-2 Concept of Physics for Class-12

(Page-230)

**Question-2 :-**

If a charged particle kept at rest experiences an electromagnetic force,

(a) there must be an electric field

(b) there must be a magnetic field

(c) both fields cannot be zero

(d) both fields can be non-zero

**Answer-2 :-**

The options **(a) and (d)** are correct

**Explanation:**

As the charged particle is at rest, its velocity,

V = 0 and magnetic force,

F = qVB = 0.

Hence,

we cannot determine whether a magnetic field is present or not. But as the particle at rest experiences an electromagnetic force, the electric field must be non-zero. As electric force acts on a particle, whether it is at rest or in motion, an electric force must be present.

**Question-3 :-**

If a charged particle projected in a gravity-free room deflects,

(a) there must be an electric field

(b) there must be a magnetic field

(c) both fields cannot be zero

(d) both fields can be non-zero

**Answer-3 :-**

The options **(c) and (d)** are correct

**Explanation:**

As the particle gets deflected, a force acts on the particle.

So, either it has got deflected due to the magnetic force or electric force;

so, both the fields cannot be zero. Also, the particle can be deflected under the combined effect of magnetic and electric forces; so, both fields can be non-zero.

**Question-4 :-**

A charged particle moves in a gravity-free space without change in velocity. Which of the following is/are possible?

(a) *E* = 0, *B* = 0

(b) *E* = 0, *B* ≠ 0

(c) *E* ≠ 0, *B* = 0

(d) *E* ≠ 0, *B* ≠ 0

**Answer-4 :-**

The options **(a), (b) and (d)** are correct

**Explanation:**

A charged particle can move in a gravity-free space without any change in velocity in the following three ways:

**(1) **

E = 0, B = 0, i.e. no force is acting on the particle and hence, it moves with a constant velocity.

**(2)**

E = 0, B ≠ 0. If magnetic field is along the direction of the velocity v, then the force acting on the charged particle will be zero,

as F = q v × B = 0. Hence, the particle will not accelerate.

**(3)**

If the force due to magnetic field and the force due to electric field counterbalance each other, then the net force acting on the particle will be zero and hence, the particle will move with a constant velocity.

**Question-5 :-**

A charged particle moves along a circle under the action of possible constant electric and magnetic fields. Which of the following is possible?

(a) *E* = 0, *B* = 0

(b) *E* = 0, *B* ≠ 0

(c) *E* ≠ 0, *B* = 0

(d) *E* ≠ 0, *B* ≠ 0

**Answer-5 :-**

The options **(b) **are correct

**Explanation:**

The electric field exerts a force *q E* on the charged particle, which always accelerates (increases the speed) the particle. The particle can never be rotated in a circle by the electric field because then the radius of the orbit will keep on increasing due to the acceleration, which is not possible.

**So, options (c) and (d) are incorrect. **

On the other hand, a magnetic field does not change the magnitude of the velocity but changes only the direction of the velocity. Since the particle is moving in a circle, where its speed remains constant and only the direction of velocity changes, so it can only be achieved

** if E = 0 and B ≠ 0.**

**Question-6 :- (Magnetic Field Obj-2 HC Verma )**

**A charged particle goes undeflected in a region containing an electric and a magnetic field. It is possible that
**

A charged particle goes undeflected in a region containing an electric and a magnetic field. It is possible that

(a) E¯ || B¯, v¯ || E¯

(b) E¯ is not parallel B¯

(c) v¯ || B¯ but v¯ is not parallel

(d) E¯ || B¯ but v¯ is not parallel

**Answer-6 :-**

The options **(a) and (b) **are correct

**Explanation:**

In option (a) velocity, electric field and magnetic field are parallel to each other.

So, the particle may accelerate but always continue to travel in the same straight path or go undeflected.

Another possibility of the particle to go undeviated is that magnetic force acting on it is counterbalanced by electric force. This is possible if all the three,

ex. velocity, magnetic field and electric field are perpendicular to each other, so that magnetic force is balanced by electric force.

So option (b) can also be one possibility. But (c) and (d) are wrong statements.

**Question-7 :-**

If a charged particle moves unaccelerated in a region containing electric and magnetic fields

(a) E‾ must be perpendicular → B‾

(b) v‾ must be perpendicular → E‾

(c) must be perpendicular to v_ B

(d) must be perpendicular to v_ B

**Answer-7 :-**

The options **(a) and (b) **are correct

**Explanation:**

Charged particle is not accelerated, the field E‾ cannot be parallel to velocity .

Hence,

the velocity is perpendicular to the electric field E‾.

The magnetic force, i.e. force due to the magnetic field, acts in a direction perpendicular to the plane containing V‾ and B‾. Hence, in order to counterbalance the magnetic force, an equal and opposite electric force must be applied along the same axis in which the magnetic force is acting.

Hence must be perpendicular to v‾ and B.

**Question-8 :-**

Two ions have equal masses but one is singly-ionised and the other is doubly-ionised. They are projected from the same place in a uniform magnetic field with the same velocity perpendicular to the field.

(a) Both ions will move along circles of equal radii.

(b) The circle described by the singly-ionised charge will have a radius that is double that of the other circle.

(c) The two circles do not touch each other.

(d) The two circles touch each other.

**Answer-8 :-**

The options **(b) and (d) **are correct

**Explanation:**

The radius of the orbit of a charged particle in an external magnetic field,

where *r* is the radius of the circle, *m* is the mass of the ion, *V* is the velocity with which the ion is projected, *q* is the charge on the ion and *B* is the uniform magnetic field.

Since the mass *m*, the velocity *V* and the magnetic field *B* are same for both the ions, *r* is inversely proportional to the charge on the ion.

Hence,

the radius of the circle described by the singly-charged ion will be twice the radius of the circle described by doubly-ionised ion.

**Question-9 :-**

An electron is moving along the positive *x*-axis. You want to apply a magnetic field for a short time so that the electron may reverse its direction and move parallel to the negative *x*-axis. This can be done by applying the magnetic field along

(a) *y*-axis

(b) *z*-axis

(c) *y*-axis only

(d) *z*-axis only

**Answer-9 :-**

The options **(a) and (b) **are correct

**Explanation:**

Any magnetic field, except one parallel to the direction of velocity can change the direction of the particle.

Hence,

Either the magnetic field along y-axis or along z-axis can reverse the direction of the particle, as the velocity is along the *x* direction.

**Question-10 :- (Magnetic Field Obj-2 HC Verma )**

Let and E‾ and B‾ denote electric and magnetic fields in a frame *S* and E‾→ and B‾ in another frame *S*‘ moving with respect to *S* at a velocity V‾ Two of the following equations are wrong. Identify them.

**Answer-10 :-**

The options **(a) and (c) **are correct

**Explanation:**

Electric force due to a charged particle is q E and magnetic force is q V B.

We can sort out the two wrong equations using dimensional analysis.

**Now, equating the above two forces. we get:**

**E = V B**

Hence,

Analysing the answers using dimensional analysis, we see that the second term on the RHS of the equations (b) and (c) are not dimensionally correct.

Thus, the options (b) and (c) are wrong.

—: End of Magnetic Field **Obj-2 HC Verma (MCQ-2) ** Solutions Vol-2 Chapter-34 :–

Return to — HC Verma Solutions Vol-2 Concept of Physics

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