FOLLOW US ON TWITTER
SHARE THIS PAGE ON FACEBOOK, TWITTER, WHATSAPP ... USING THE BUTTONS ON THE LEFT


YOUR PARTICIPATION FOR THE GROWTH OF PHYSICS REFERENCE BLOG

Thursday, January 2, 2020

A long solenoid has an area of cross-section of 28 cm2, as shown in Fig. 5.1.


Question 7
(a) Define the tesla. [2]


(b) A long solenoid has an area of cross-section of 28 cm2, as shown in Fig. 5.1.


Fig. 5.1

A coil C consisting of 160 turns of insulated wire is wound tightly around the centre of
the solenoid.

The magnetic flux density B at the centre of the solenoid is given by the expression
B = μ0n I

where I is the current in the solenoid, n is a constant equal to 1.5 × 103 m-1 and μ0 is
the permeability of free space.

Calculate, for a current of 3.5 A in the solenoid,

(i) the magnetic flux density at the centre of the solenoid, [2]

(ii) the flux linkage in the coil C. [2]


(c) (i) State Faraday’s law of electromagnetic induction. [2]


(ii) The current in the solenoid in (b) is reversed in direction in a time of 0.80 s.
Calculate the average e.m.f. induced in coil C. [2]





Reference: Past Exam Paper – June 2013 Paper 41 & 43 Q5





Solution:
(a) The tesla is the (uniform magnetic) flux is normal to a long (straight) wire carrying a current of 1 A, creates a force per unit length of 1 N m–1.



(b)
(i)
{Magnetic flux density B = μ0n I}
Magnetic flux density B = (4Ï€×10–7) × (1.5×103) × 3.5
Magnetic flux density B = 6.6 × 10–3 T

(ii)
{Flux linkage = Magnetic Flux × Number of turns
Flux linkage = BA × N
Cross-sectional area A should be in SI units.}

Flux linkage = (6.6 × 10–3) × (28 × 10–4) × 160
Flux linkage = 3.0 × 10–3 Wb


(c) (i) Faraday’s law of electromagnetic induction states that the (induced) e.m.f. is proportional to rate of change of (magnetic) flux (linkage).

(ii)
{To reverse the direction of the current, the current must first be reduced to zero, and then increased to the same value in the opposite direction. So, the flux linkage calculated above should be  considered twice.

Average e.m.f. induced = change in flux linkage / time}

Average e.m.f. = (2 × 3.0 × 10–3) / 0.80
Average e.m.f. = 7.4 × 10–3 V

4 comments:

  1. Why is the area of the solenoid used to determine the flux linkage in the coil in equation phi = BAN? Please explain!

    ReplyDelete
    Replies
    1. for one turn (circle), flux density = BA
      the area of the coil is used

      A solenoid consists of several turns. We need to account for this by using the number of turns, N.
      For a solenoid, we use magnetic flux linkage instead of flux density.

      Delete
    2. But we are using the number of turns of the coil, not the solenoid.

      Why do we use the area of the solenoid again? Since determining flux linkage in the coil should be done by the equation flux linkage in coil = B through coil (which is same as B at the centre of solenoid) × area of coil × number of turns in coil.

      Delete
    3. Note that the explanation has been updated slightly.

      Magnetic flux density B is obtained from the formula in the question.

      Flux linkage = Magnetic flux x number of turns
      Magnetic flux = BA
      Flux linkage = BA x N

      note also that the area A has NOT been used when calculating the magnetic flux density B (using formula in question)

      Delete

If it's a past exam question, do not include links to the paper. Only the reference.
Comments will only be published after moderation

Currently Viewing: Physics Reference | A long solenoid has an area of cross-section of 28 cm2, as shown in Fig. 5.1.