# Magnetism Exam 1 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**
**Bresultant=B₁+B₂+B₃**

**Bresultant=√(12.10⁻⁶-16.10⁻⁶)²+(8.10⁻⁶)²**

**Bresultant=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;

BA: inward

BB:inward

BC:outward

**BY=BA+BB-BC**

**BY=2k.I/d+2k.3I/d-2k.I/d=2k.3I/d**

Ratio of magnetic fields;

**BX/BY=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.