Question

A charged particle +q in an electric field `vecE` experiences a force in the direction of the electric field. As a result, its kinetic energy changes. Similarly, the charged particle also experiences a force when it moves in a magnetic field `vecB`. But this magnetic force is perpendicular to both the velocity `vecv` of the charged particle and the magnetic field `vecB`, so it cannot change the kinetic energy of the charged particle. Consider two charged particles 1 and 2 of masses m and `m/2` having charges −q and +2q, respectively. They are accelerated from rest through the same potential difference V and acquire kinetic energy K1 and K2. Then they enter in a region of uniform magnetic field `vecB` perpendicular to their velocities.

(i) The ratio of their kinetic energies `((K_1)/(K_2))` is:   [1]

1. `1/2`
2. `1/4`
3. 4
4. 1
5.  

(ii) The ratio of the radii of the circular paths described by them `((r_1)/(r_2))` is:   [1]

1. `1/sqrt2`
2. `sqrt2`
3. `1/2`
4. 2

(iii) Suppose particles 1 and 2 enter the magnetic field `vecB = B_0hatk` with velocities `vecv_1 = v_1hati and vec2 = v_2hati`. Then:   [1]

1. Both particles revolve clockwise
2. Both particles revolve anticlockwise
3. Particle 1 revolves clockwise while particle 2 revolves anticlockwise.
4. Particle 1 revolves anticlockwise while particle 2 revolves clockwise.

(iv) (a) If period of revolution for particle 1 is 4 s, then for particle 2, the period will be:   [1]

1. 1 s
2. 2 s
3. 4 s
4. 8 s

OR

(iv) (b) If the value of momentum for particles 1 and 2 are p₁ and 1 p₂, then:   [1]

1. `p_1 = (p_2)/(2)`
2. p₁ = p₂
3. `p_1 = 2p`
4. `p_1 = 4p_2`

Answer 1

(i) `1/2`

Explanation:

Kinetic energy gained:

K = qV

K₁ = 1qV   ...(i)

K₂ = 2qV   ...(i)

Equation (i) and (ii) we get,

`((K_1)/(K_2)) = (qV)/(2qV)`

= `1/2`

(ii) `1/sqrt2`

Explanation:

Radius in magnetic field:

r = `(mv)/(qB)`

Using K = `1/2 mv^2`

v = `sqrt((2K)/m)`

r = `m/(qB)sqrt((2K)/m)`

= `sqrt((2mK)/(qB))`

K = qV

r α `sqrt(m)/q`

`(r_1)/(r_2) = sqrt((m_1  q_2)/(m_2  q_1))`

= `sqrt((m  .  2q)/((m//2)  .  q))`

= `sqrt((2m)/(m//2))`

= `sqrt4`

= `sqrt2`

`(r_1)/(r_2) = 1/sqrt2`

(iii) Particle 1 clockwise, particle 2 anticlockwise

Explanation:

Magnetic force: `vecF = q(vecv xx vecB)`

Since the charges are of opposite signs, the direction of the magnetic force differs for each particle. 

A negative charge experiences force in the direction opposite to that given by the right-hand rule, whereas a positive charge follows the right-hand rule.

Therefore, the two particles move in opposite directions.

(iv) (a) 1 s

Explanation:

Time period in magnetic field:

T = `(2pim)/(qB)`

`T α m/q`

= `m/q`

`(T_1)/(T_2) = 4`

T₁ = 4 s    ...(Given)

T₂ = 1 s
OR

(iv) (b) p₁ = p₂

Explanation:

Momentum: p = mv

Using K = `1/2  m v^2`

= qV

p = `sqrt(2mK)`

Since K ∝ q,

p ∝ `sqrt(mq)`

`p_1 = sqrt(2m  .  qV)`   ...(i)

`p_2 = sqrt(m/2  .  2qV)`   ...(i)

Equations (i) and (ii) we get,

`(p_1)/(p_2) = 1`

∴ p₁ = p₂