**Magnetism Exam1 ****and Problem Solutions**

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

**F=k.M _{1}.M_{2}/r^{2}=(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 I_{1}, I_{2} and I_{3}.

**Magnitudes of magnetic fields;**

**B _{1}=2k.6/0,1=12.10^{-7}/10^{-1}=12.10^{-6 }N/Amps.m**

**B _{2}=2k.4/0,1=8.10^{-7}/10^{-1}=8.10^{-6 }N/Amps.m**

**B _{3}=2k.8/0,1=16.10^{-7}/10^{-1}=16.10^{-6 }N/Amps.m**

**B _{resultant}=B_{1}+B_{2}+B_{3}**

**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.**

**B _{A}: outward**

**B _{B}:inward**

**B _{C}:inward**

**B _{X}=B_{B}+B_{C}-B_{A}**

**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=Φ _{2}-Φ_{1}**

**Φ _{1}=0, since cross section area of solenoid and magnetic field lines are parallel to each other.**

**Φ _{2}=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.

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