# Questions OBJECTIVE - II and Answer Magnetic Field HC Verma Part II

Q#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: (a), (d).

There will not be a magnetic force on a charged particle at rest even if there is a magnetic field present. Option (d) is correct not option (b).

But if there is an electric field, the charged particle can not remain at rest in absence of a magnetic force on it. So the electric field here must be zero. Option (a) is correct and option (c) is not correct.

Q#2.

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

(a) the electric field must not be zero.

(b) the magnetic field must not be zero.

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

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

Answer: (a), (d).

A magnetic field does not exert a force on a charged particle kept at rest. So if a charged particle is at rest, there may or may not exist a magnetic field. Even if the particle experiences an electromagnetic force, there must be the presence of an electric field. Hence only options (a) and (d) are correct.

Q#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 nonzero.

Answer: (c), (d).

If a charged particle projected in a gravity-free room deflects, it may be due to either of the two fields. Hence options (a) and (b) are not correct.

If both fields are zero, then force on the particle is zero and it can not deflect. So both fields can not be zero. Option (c) is correct.

If both fields are present with nonzero magnitudes, even then the particle can deflect. Option (d) is correct.

Q#4.

A charged particle moves in a gravity-free space without a 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: (a), (b), (d).

If both fields are zero, there will be no force on the charged particle. Thus, it can move without a change in velocity. Option (a) is correct.

If there is only the magnetic field present, even then the charged particle can move with a constant velocity if the direction of the magnetic field is parallel to the direction of the velocity. Option (b) is correct.

If the only electric field is present then the charged particle will always experience a force due to it and it can not move with a constant velocity. Option (c) is not correct.

The magnetic force on a moving charged particle is perpendicular to the magnetic field and the velocity. Now if the direction of the electric field is such that it exerts a force equal and opposite to the magnetic force, then the resultant force on the particle is zero and it can move with a constant velocity. Option (d) is correct. See the picture below:-

Q#5.

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

(a) E = 0, B =0.

(b) E = 0, B ≠ 0.

(c) E ≠ 0, B = 0.

(d) E ≠ 0, B ≠ 0.

Answer: (b).

Without a centripetal force a particle can not move along a circle. In the absence of both the fields, the force on the particle is zero. So both fields can not be zero in this case. Option (a) is incorrect.

If the electric field is zero but the magnetic field is non-zero and the charge moves perpendicular to the magnetic field, a magnetic force perpendicular to the velocity will act. It will provide a centripetal force to the particle and it will move in a circle in a plane perpendicular to the magnetic field. Option (b) is correct.

If the magnetic field is zero but the electric field is non zero then the charged particle can not move along a circle because it can not provide a constant magnitude of force always perpendicular to the velocity in any condition. Option (c) is incorrect.

If both the fields are non-zero, the charged particle will have an additional straight-line velocity and due to the combined effects of these velocities, the particle cannot move in a complete circle. Option (d) is incorrect.

Q#6.

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

(a) E ॥ B, v ॥ E

(b) E is not parallel to B

(c) v ॥ B but E is not parallel to B.

(d) E ॥ B but v is not parallel to E.

Answer: (a),(b).

If the direction of velocity, electric and magnetic fields are all parallel, the magnetic force will be zero on the particle. Only the electric force will act parallel to the velocity. It will move along a straight line without deflection. Option (a) is correct.

If the particle moves in a magnetic field, not in a direction parallel to this field, a magnetic force perpendicular to both the velocity and the magnetic field will act on the particle. If an electric field is present such that it applies an electric force that is equal and opposite to the magnetic force on the charge then the resultant force on the particle is zero. The charge will move in a straight line undeflected. Option (b) is correct.

If the direction of velocity is parallel to the magnetic force, no magnetic force is on the charged particle. Now if the direction of the electric field is not parallel to the magnetic force/velocity then there will be an electric force on the particle that is not along the line of velocity, hence it will deflect. Option (c) is not possible.

In the last case, since the velocity is not parallel to the magnetic field, a magnetic force will act on it that is perpendicular to both B and v. It can move undeflected only if the electric field is along the magnetic force i.e. perpendicular to B but given that E || B. So in this situation the condition cannot be fulfilled. So, the option (d) is not possible.

Q#7.

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

(a) E must be perpendicular to B.

(b) v must be perpendicular to E.

(c) v must be perpendicular to B.

(d) E must be equal to vB.

Answer: (a), (b).

A particle can move unaccelerated only when the resultant force acting on it is zero. If there are both electric and magnetic fields present in a region, only in the following situation resultant force will be zero.

*q*|**E**|
= q|**v** x **B**| (i)

So, the line of action of E is the same as the line of action of vxB, that is E ⟂ B and v ⟂ E. Options (a) and (b) are correct.

From (i), it is clear that there is no necessity of v being perpendicular to B. Option (c) is incorrect.

Also, from (i), E = vB.sinθ, and since θ is not necessarily 90°, so option (d) is not correct.

Q#8.

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

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

(b) The circles described by the single ionized charge will have a radius double that of the other circle.

(c) The circles do not touch each other.

(d) The two circles touch each other.

Answer: (b), (d).

Centripetal force,

*qvB* = *mv*²/*r*

*r* = *mv*/*qB*

m, v and B are constant here, so the radius of the circle for 2q charged ion will be r/2. Thus the circle described by the singly charged ion will have a radius double the doubly charged ion. Option (b) is correct, option (a) is wrong.

Since both the ions are projected from the same place, the two circles will touch each other. Option (d) is correct, option (c) is wrong

Q#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: (a), (b).

Since the force exerted due to a magnetic field is perpendicular to the velocity and also perpendicular to the field itself, a magnetic field applied along Y-axis will move the electron on a circular path in the XZ plane. If applied along the Z-axis, the electron will move on a circular path in the XY plane. In both of these cases, the magnetic field may be withdrawn just after the electron has moved half-circle, resulting in reversal of the movement of the electron along parallel to the negative X-axis. So options (a) and (b) are correct, other two are wrong.

Q#10.

Let 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.

(a) Bᵧ' = Bᵧ + $\frac{vEz}{c^2}$

(b) Eᵧ' = Eᵧ + $\frac{vBz}{c^2}$

(c) Bᵧ' = Bᵧ + *vEz*

(d) Eᵧ' = Eᵧ + *vBz*.

Answer: (b), (c).

Since the magnitude of electric force =qE and that of magnetic force =qvBsinθ. In case they are equal,

*E* = *vB* sinθ.

In case that θ = 90°, sinθ =1. Then,

E = *vB* and *B*
= *E*/*v*

So, the dimensions of E are the same as vB, and the dimensions of B are the same as E/v. The dimensions of the last term of the R.H.S. of the second equation do not match with E. This equation is wrong. Option (b) is correct.

The same is the case of the third equation. The last term does not dimensionally match with B. This equation is also incorrect. So, option (c) is true.

The first and the last equations are dimensionally correct. Since it is given that two equations are wrong that we have found out as (b) and (c). There is no necessity to check the facts of the rest two equations.

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