Magnetism Exam1 and  Problem Solutions

1. Find the forces exerted by S poles of magnets given below.

F=k.M₁.M₂/r²=(10^-7.10^-4.10^-3)/(0,6)²

F=10^-14/(36.10^-2)

F=10^-12/36

2. Find resultant magnetic field at point O, produced by I₁, I₂ and I₃.

Magnitudes of magnetic fields;

B₁=2k.6/0,1=12.10^-7/10^-1=12.10^-6 N/Amps.m

B₂=2k.4/0,1=8.10^-7/10^-1=8.10^-6 N/Amps.m

B₃=2k.8/0,1=16.10^-7/10^-1=16.10^-6 N/Amps.m

B_resultant=B₁+B₂+B₃

B_resultant=√(12.10^-6-16.10^-6)²+(8.10^-6)²

B_resultant=4√5.10^-6 N/Amps.m

3. A, B and C wires are given below. Find the magnetic field of A, B and C at points X and Y.

Directions of magnetic fields at point X are found using right hand rule.

BA: outward

BB:inward

BC:inward

BX=BB+BC-BA

BX=2k.3I/3d+2k.I/d-2k.I/d=2k.I/d

Directions of magnetic fields at point Y are;

B_A: inward

B_B:inward

B_C:outward

 B_Y=B_A+B_B-B_C

B_Y=2k.I/d+2k.3I/d-2k.I/d=2k.3I/d

Ratio of magnetic fields;

B_X/B_Y=2k.I/d/2k.3I/d=1/3

4. Solenoid having number of loops N and surface area A is shown in picture given below. If we change the position of solenoid as shown in the picture below, find the equation used for finding induced emf of solenoid.

Induced emf=ε=-(∆Φ)/(∆t).N

Change in Flux;

∆Φ=Φ₂-Φ₁

Φ₁=0, since cross section area of solenoid and magnetic field lines are parallel to each other.

Φ₂=B.A

∆Φ=B.A-0=B.A

ε=-B.A.N/t

5. Draw the directions of magnetic field lines at point A, B, C and D in the picture given below.

Directions of magnetic field lines are drawn from N pole to S pole as shown in the picture given below.