मराठी

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.

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प्रश्न

Read the following paragraphs and answer the questions that follow:

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

(ii) The ratio of the radii of the circular paths described by them `((r_1)/(r_2))` is ______.

  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. 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 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 p2, then ______.

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

(i) The ratio of their kinetic energies `((K_1)/(K_2))` is`bbunderline(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) The ratio of the radii of the circular paths described by them `((r_1)/(r_2))` is `bbunderline(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 ∝ `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) Suppose particles 1 and 2 enter the magnetic field `vecB = B_0hatk` with velocities `vecv_1 = v_1hati and vec2 = v_2hati`. Then 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) If period of revolution for particle 1 is 4 s, then for particle 2, the period will be 1s.

Explanation:

Given: T1 = 4 s

Mass m1 = m

q1 = q

 Mass m2 = `m/2`

q2 = 2q

Time period in magnetic field:

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

`T α m/q`

`(T_1)/(T_2) = ((m_1/q_1))/((m_2/q_2))`

Substitute the given values into the ratio:

`(T_1)/(T_2) = ((m/q))/(((m/2)/(2q)))`

`(T_1)/(T_2) = (m/q)/(m/(4q))`

`(T_1)/(T_2) = m/q xx (4q)/m`

`(T_1)/(T_2) = 4`

`4/(T_2) = 4`

`T_2 = 4/4`

T2 = 1 s

OR

(iv) (b) If the values of momentum for particles 1 and 2 are p1 and p2, then `bbunderline(p_1 = p_2)`.

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)`   ...(ii)

From equations (i) and (ii) we get,

`(p_1)/(p_2) = 1`

∴ p1 = p2

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