Hello friends, I hope you all are doing great. In today’s tutorial, we will have a look at 2nd Year Physics Chapter 14 Solved Questions. I have started a series of tutorials related to the solution of questions given in 2nd-year physics. In the previous tutorial, I have discussed all the questions of chapter 13 with the detailed.

In this post, we will have a detailed look at all the questions given in chapter 14 of FSC physics. I will explain all the questions one by one and give an answer to every question in a simple way. So let’s get started with 2nd Year Physics Chapter 14 Exercise Solved Questions.

#### 2nd Year Physics Chapter 14 Exercise Solved Questions

• So friends let’s discuss all the questions given in the exercise of chapter 14 with detail.

#### Question 14.1

• For extreme (maximum) flux, the direction of a surface of a plane should be at 90 degrees to the direction of a magnetic field or vector area is parallel to the field. As we know that.

φ = (B) . (A)

φ = BAcosθ

φ = BAcos0

φ = BA

• In this case, the maximum flux will be across the loop.
• If the surface of the conductive loop is parallel or vector area is at 90 degrees to the field line direction the flux will be maximum.

φ = (B) • (A)

φ = BAcosθ

φ = BAcos90

• In this case, the flux will be zero or minimum.

#### Question 14.2

For Stationary Charges

• The static or stationary charges have no ability to generates a magnetic field but they generate an electric field around it.

For Moving Charges

• Moving charges generate a magnetic field about their movement path like the field produced in the current-carrying conductor.

#### Question 14.3

• As we know the formula of magnetic field for the current-carrying solenoid is given here.

B=μonl

since n=N/L

so, B=μo(Nl)/L

(a):

• With the double-length we have.

B=μo(Nl)/2L

B=1/2.μo(Nl)/L

B=1/2.B

• With double of length field reduces to half.

(b):

• If we double the number of turns.

B’=(μo 2NI) /L

B’=2 (μoNI) /L

B’=2B

• With the doubling, the number of turns field will also double.

#### Question 14.4

• The formula for the force applying on the proton is given here.

FB=q(VxB)

• When the proton is moving in a positive X-axis magnetic field is along the negative Z-axis.
• If we use right-hand rule than we will find that the direction of the magnetic force will be in the positive Y-axis.
• Due to the magnetic force, the direction of motion of electron will not be along the X-axis but electron move in a circular path in the X-Y plane.

#### Question 14.5

• If 2 charged particles are in a region where the field is at 90 degrees to their velocities both charges will deflect in the opposite direction.
• It shows that charges of these particles are opposite means one is a positive charge and the other is a negative charge. Also can say that one is an electron and the other is a proton.

#### Question 14.6

• The force on a charged particle is always at 90 degrees to its movement direction.
• Due to the force particle will change its movement direction to the circular shape.
• The work done will be given as.

W=F.d

• So the angle among F and d is 90 degrees. W=F.d

W=Fdcosθ

W=Fdcos90

W =0

• There will be zero work done by the field.

#### Question 14.7

• No the field for this region will not be zero since charge particle can move in a straight path when the force on the charge is ‘0’.
• There are 2 situations here.

(1):

• If direction of B and V are the same means θ = 0.
• As we know that.

F=qvBsinθ

F=qvBsin0

F=0

(2):

• If V and B are not in the same direction or opposite direction. For this condition θ=180.

F = qvBSinθ

F=qvBsin180

F=0

#### Question 14.8

• The photo displaying on the screen of the television is created through the beam of electrons.
• If we place magnet in front of TV screen it will disturb the path of electrons displaying the picture on the screen. So image created through the beam of electrons will be deformed.

#### Question 14.9

• The current-carrying loop placing in a filed will bear a torque given here.

T=IBAcosα

• If the plane of the loop is at 90 degrees to field.

T=IBAcos90

t=0

• For this condition, there will be no movement in the loop.

#### Question 14.10

• Torque applying on the current-carrying loop in a magnetic field is given here.

t=NlBAcosα

• So if we put the current loop in the specific region if it deflected from its place there will be a field exits in that region.

#### Question 14.11

• For separation of isotopes, the ions of elements are passed through the uniform filed positioned at 90 degrees.
• The magnetic force will act on them.
• This force on ions rotates them at different circular paths of radii.

qvB=mv2/r

r=mv/qB

• q and B are constant.

r α m

• So the isotopes of a large mass will be moved at a larger circular path radius than other isotopes.
• So all isotopes are at the different positions according to their radius by using Mass spectrometer they are separated.

#### Question 14.12

(a):

• If the plane of a current-carrying loop is parallel to the field then the torque induces will be maximum.

t=NlABcosα

t=NlABcos0  since  cos0=1

t=NlAB

(b):

• If the plane of the current-carrying coil is at 90 degrees to magnetic field torque induced will be minimum.

t=NlABcosα

t=NlABcos90  since  cos90=0

t=0

• As ammeter is always connected in series to measure the current, therefore it should have very small resistance. If an ammeter has high resistance then it will alter the current of the circuit.

#### Question 14.13

• For both, AC and DC currents loop will rotate.
• Because for both the currents loop plane and magnetic field are at 90 degrees to each other. So induced torque will be ‘0’.

t=NlABcosα

t=NlABcos90  since  cos90=0

t=0

#### Question 14.14

• For the measurement of a current of any circuitry, ammeter is connected in series. So it should provide less value resistance.
• If the resistance of ammeter is large it will operate as a load and due to large resistance, some losses will occur across this resistance.
• This reduces the current from the circuitry and current will not be calculated correctly.

#### Question 14.15

• For voltage measurements voltmeter is connected in parallel across the circuit. If its resistance is low it will damage and not measure voltage correctly.\

#### Related Posts

So, friends, it is the detailed post about the 2nd Year Physics Chapter 14 Exercise Solved Questions if you have any further query ask in comments. see you in next post have a good day. Thanks for reading.

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