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

Q#1

A positively charged particle projected towards the east is deflected towards the north by a magnetic field. The field maybe

(a) towards west

(b) towards south

(c) upwards

(d) downwards

Answer: (d)

Since the deflection of the positively charged particle is towards the north, the magnetic field causing this deflection may be perpendicular to both directions i.e. in the vertical direction.

The force by the magnetic field is given as,

*F* = $qv \times B$

From this vector cross-product relation, we can conclude that the direction of the magnetic field is downwards. Option (d) is correct.

Q#2

A charged particle is whirled in a horizontal circle on a frictionless table by attaching it to a string fixed at one point. If a magnetic field is switched on in the vertical direction, the tension in the string

(a) will increase

(b) will decrease

(c) will remain the same

(d) may increase or decrease.

Answer: (d)

Neither the direction of the movement of the particle is known (clockwise or anticlockwise) nor the direction of the magnetic field is given (upwards or downwards). The nature of the charge on the particle is also not given. Depending on the nature of the charge and these two directions, the force on the charged particle may be either towards the fixed point (center) or away from it. So the tension in the string may increase or decrease. Option (d) is correct.

Q#3.

Which of the following particles will experience a maximum magnetic force (magnitude) when projected with the same velocity perpendicular to a magnetic field?

(a) electron

(b) proton

(c) He⁺

(d) Li⁺⁺

Answer: (d)

The maximum magnitude of the force is

*F* = *qvB *

Given that v and B are fixed, hence the particle having maximum charge will experience the maximum magnetic force. Out of the four particles, electron, proton and He⁺ have the same magnitude of charge but Li⁺⁺ has double that magnitude. Hence Li⁺⁺ will experience the maximum magnetic force. Option (d) is correct.

Q#4.

Which of the following particles will describe the smallest circle when projected with the same velocity perpendicular to a magnetic field?

(a) electron

(b) proton

(c) He⁺

(d) Li⁺

Answer: (a)

The magnitude of the charge on all of the given particles is the same. Hence the magnetic force F will have equal magnitudes. This magnetic force provides the centripetal force for the circular motion of the particles, hence

*F* = *mv*²/*r*

*r* = *mv*²/*F*

v²/F is fixed, so the radius r will be minimum for the particle having the least mass. Here it is the electron. Thus option (a) is correct.

Q#5. Which of the following particles will have a minimum frequency of revolution when projected with the same velocity perpendicular to a magnetic field?

(a) electron

(b) proton

(c) He⁺

(d) Li⁺

Answer: (d)

The time period T of a circular motion of a particle with speed v is,

*T* = $\frac{2\pi r}{v}$

So the frequency,

*f* = $\frac{1}{T}$ = $\frac{v}{2\pi r}$

Also the magnetic force,

*qvB* = *mv*²/*r*

$v=\frac{qBr}{m}$

Now frequency,

$f=\frac{qBr}{2 \pi r m}=\frac{qB}{2\pi m}$

Here for the given particles, the magnitude of charge q is the same, B is constant, thus the frequency f is inversely proportional to the mass m of the particle. Since Li has the greatest mass among the four particles, it will have the minimum frequency. Option (d) is correct.

Q#6.

A circular loop of area 1 cm², carrying a current of 10 A, is placed in a magnetic field of 0.1 T perpendicular to the plane of the loop. The torque on the loop due to the magnetic field is

(a) zero

(b) 10⁻⁴ N.m

(c) 10⁻² N.m

(d) 1 N.m

Answer: (a)

Since the direction of the magnetic field is perpendicular to the plane of the loop, the torque on the loop will always be zero, whatever be the current in the wire, the area of the loop, or the magnitude of the magnetic field. Option (a) is correct.

Q#7.

A beam consisting of protons and electrons moving at the same speed goes through a thin region in which there is a magnetic field perpendicular to the beam. The protons and electrons

(a) will go un-deviated

(b) will be deviated by the same angle and will not separate

(c) will be deviated by different angles and hence separate

(d) will be deviated by the same angle but will separate.

Answer: (c)

The magnitude of the charge on a proton or an electron is the same. Hence the force on them in the thin magnetic region will be the same but opposite in direction because the nature of the charge on them is opposite.

Due to the force the acceleration in the direction of the force = F/m. Mass is different for them hence the different accelerations. Otherwise, we may think that in the thin magnetic field, the radius of the arc on which they move = mv²/F.

Due to the difference in the mass, they will deviate by different angles and separate. Option (c) is correct.

Q#8.

A charged particle in a uniform magnetic field. The velocity of the particle at some instant makes an acute angle with the magnetic field. The path of the particle will be

(a) a straight line

(b) a circle

(c) helix with uniform pitch

(d) a helix with a non-uniform pitch.

Answer: (c)

We can resolve the velocity along the magnetic field and perpendicular to the magnetic field. Due to the perpendicular component of the velocity, the charged particle will move in a circle in a plane perpendicular to the direction of the magnetic field. On the other hand, the component of the velocity along the magnetic field will be unaffected and due to this, the particle will have a uniform straight motion along the field. When we combine these two motions for the charged particle, the path will be a helix with a uniform pitch. Option (c) is correct.

Q#9.

A particle moves in a region having a uniform magnetic field and a parallel, uniform electric field. At some instant, the velocity of the particle is perpendicular to the field direction. The path of the particle will be

(a) a straight line

(b) a circle

(c) a helix with uniform pitch

(d) a helix with a non-uniform pitch.

Answer: (d)

The effect of the magnetic field will be to move the charged particle in a circular path while the effect of the electric field on the particle will be to accelerate it towards either of the directions of the field depending upon the nature of the charge. So combining a uniform circular motion and an accelerated straight motion perpendicular to the circular motion will give a path of a helix with a non-uniform pitch. Option (d) is correct.

Q#10.

An electric current 'i' enters and leaves a uniform circular wire of radius 'a' through diametrically opposite points. A charged particle q moving along the axis of the circular wire passes through its center at speed v. The magnetic field acting on the particle when it passes through the center has a magnitude

(a) $qv \left(\frac{\mu_0 i}{2a}\right)$

(b) $qv \left(\frac{\mu_0 i}{2\pi a}\right)$

(c) $qv \left(\frac{\mu_0 i}{a}\right)$

(d) *zero*.

Answer: (d)

Since the current 'i' enters and leaves the wire loop at diametrically opposite points, the magnitude of the current in each half of the wire will be equal to i/2. The direction of the current in one half will be clockwise while in the other it will be anti-clockwise. So whatever magnetic field they produce at the center will be equal and opposite in direction. Thus the net magnetic field at the center = zero. Option (d) is correct.

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