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Q1. A proton and an α particle having the same kinetic energy are in turn allowed to pas through a uniform magnetic field perpendicular to the direction of motion. The ratio of the radii of their path is

• 1) 1:2
• 2) 1:4
• 3)
• 4) 1:1

Solution

Given:

Q2. An electron after being accelerated through a potential difference of 150 V enters a uniform magnetic field of 0.0005 T perpendiculars to its direction of motion. Calculate the radius of the path described by the electron.

Solution

Here we are given,                    

Q3. A proton and an alpha particle enter a uniform magnetic field perpendicularly with the same speed. How many times is the time period of the alpha particle than that of the proton? Also find the ratio of the radii of the circular path of the two particles?

Solution

Time period of revolution is given as,   Let  m_p = mass of the proton        m_α = mass of the alpha particle.        q_p = charge of the proton        q_α = charge of the alpha particle. For proton: time period, For alpha particle: time period, => As So, Therefore, =>

Q4. A circular coil of 200 turns and radius 10 cm is placed in a uniform magnetic field of 0.5 T, normal to the plane of the coil. If the current in the coil is 3.0 A, calculate the (a) Total torque on the coil (b) Total force on the coil (c) Average force on each electron in the coil due to the magnetic field Assume the area of cross-section of the wire to be 10^-5m² and the free electron density is 10²⁹/m³.

Solution

Here n = 200, r = 10 cm = 0.1 m B = 0.5 T, I = 3 A (a) Torque is given by the relation: (b) The total force on a current loop placed in a magnetic field is always zero. (c) Given N  = 10²⁹/m³, A = 10^-5m² Average force on an electron of charge (e), moving with drift velocity (V_d) in the magnetic field (B), is given by: F = BeV_d

Q5. A toroid coil has 1000 turns and diameter of 30 cm with the area of cross section of 2 cm². Calculate i) inductance of coil ii) induced e.m.f when the a current of 2 A is reversed in 0.03 secs.

Solution

_Given: d=30 cm r=15 cm=15×10^-2 m A=2 cm² = 2×10^-4 m² di=2 A dt=0.03 s  

Q6. An electron beam was deflected in a given field which was perpendicular to the beam. The beam follows a parabolic path after deflection. Give answer whether the field was electric field or magnetic field.

Solution

The beam was deflected by an electric field. 

Q7. State Biot – Savart law. Deduce the expression for the magnetic field at a point on the axis of a current carrying circular loop of radius ‘R’ distant ‘x’ from the centre. Hence, write the magnetic field at the centre of a loop.

Solution

Q8. The electron in a hydrogen atom circles around the proton with a speed of 2.18 x 10⁶ m/s in an orbit of radius 5.3 x 10 ^-11 m. calculate the equivalent current and magnetic field produced by proton.

Solution

v = 2.18 x 106 m/s r = 5.3 x 10 -11 m Time period (T)=  =    Equivalent current I = e/T     Magnetic field produced by proton  B = 12.42T  

Q9. The velocities of two particles A and B entering a uniform electric field are in the ratio 4:1. On entering the field they move in different circular paths. Give the ratio of the radii of curvature of the paths of the particles.  OR You are given a low resistance R₁, a high resistance R₂ and a moving coil galvanometer. Suggest how would you use these to have an instrument that will be able to measure (i) current (ii) potential difference.

Solution

      OR     (i) To measure current we shall connect low resistance R₁ in parallel with the coil of moving coil galvanometer. This arrangement is called ammeter. (ii) To measure potential difference we shall connect high resistance R₂ in series with the coil of galvanometer. This arrangement is called voltmeter.

Q10. Derive an expression for the torque acting on a loop of N turns, area A, carrying current i, when held in a uniform magnetic field.

Solution

Q11. What is the length of a solenoid with total number of turns 1000, area of cross section 20 cm² and self inductance of 4.02 × 10^-3 H.

Solution

Q12. A charged particle moving in a uniform magnetic field of induction 10^-3 Wb/m²  with a velocity of 10⁶ m/s in a circular path of radius 0.45 cm. Find out the specific charge of the particle.

Solution

As we know that for circular motion of charged particle is given as,