# Chapter 4 – Moving Charges And Magnetism Questions and Answers: NCERT Solutions for Class 12 Physics

Class 12 Physics NCERT book solutions for Chapter 4 - Moving Charges And Magnetism Questions and Answers.

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Class 12 Physics NCERT book solutions for Chapter 4 - Moving Charges And Magnetism Questions and Answers.

Number of turns on the circular coil, n=100n=100

Radius of each turn, r=8.0cm=0.08mr=8.0cm=0.08m

Current flowing in the coil is given to be, I=0.4AI=0.4A

We know the expression for magnetic field at the centre of the coil as,

|B|=μ04π2πnIr|B|=μ04π2πnIr

Where, μ0=4π×10−4TmA−1μ0=4π×10−4TmA−1is the permeability of free space.

On substituting the given values we get,

|B|=4π×10−7×2π×100×0.44π×0.08|B|=4π×10−7×2π×100×0.44π×0.08

⇒|B|=3.14×10−4T⇒|B|=3.14×10−4T

Clearly, the magnitude of the magnetic field is found to be 3.14×10−4T3.14×10−4T.

Current in the wire, I=35AI=35A

Distance of the given point from the wire, r=20cm=0.2mr=20cm=0.2m

We know the expression for magnetic field as,

B=μ04π2IrB=μ04π2Ir

Where, μ0=4π×10−4TmA−1μ0=4π×10−4TmA−1is the permeability of free space.

On substituting the given values, we get,

B=4π×10−7×2×354π×0.2B=4π×10−7×2×354π×0.2

⇒B=3.5×10−5T⇒B=3.5×10−5T

Thus, we found the magnitude of the magnetic field at the given point to be 3.5×10−5T3.5×10−5T.

The current in the wire, I=50AI=50A

The distance of the given point from the wire, r=2.5mr=2.5m

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We have the expression for magnetic field as,

B=2μ0I4πrB=2μ0I4πr

Where, μ0=4π×10−4TmA−1μ0=4π×10−4TmA−1is the permeability of free space.

Substituting the given values, we get,

B=4π×10−7×2×504π×2.5B=4π×10−7×2×504π×2.5

⇒B=4×10−6T⇒B=4×10−6T

Now from Maxwell’s right hand thumb rule we have the direction of the magnetic field at the given point B to be vertically upward.

Current in the power line, I=90AI=90A

Distance of the mentioned point below the power line, r=1.5mr=1.5m

Now, we have the expression for magnetic field as,

B=2μ0I4πrB=2μ0I4πr

Where, μ0=4π×10−4TmA−1μ0=4π×10−4TmA−1is the permeability of free space.

On substituting the given values, we get,

B=4π×10−7×2×904π×1.5B=4π×10−7×2×904π×1.5

⇒B=1.2×10−5T⇒B=1.2×10−5T

We found the magnitude of the magnetic field to be 1.2×10−5T1.2×10−5Tand it will be directed towards south as per Maxwell’s right hand thumb rule.

Current in the wire, I=8AI=8A

Magnitude of the uniform magnetic field, B=0.15TB=0.15T

Angle between the wire and magnetic field, θ=30∘θ=30∘

We have the expression for magnetic force per unit length on the wire as,

F=BIsinθF=BIsinθ

Substituting the given values, we get,

F=0.15×8×1×sin30∘F=0.15×8×1×sin30∘

⇒F=0.6Nm−1⇒F=0.6Nm−1

Thus, the magnetic force per unit length on the wire is found to be 0.6Nm−10.6Nm−1

Length of the wire, l=3cm=0.03ml=3cm=0.03m

Current flowing in the wire, I=10AI=10A

Magnetic field, B=0.27TB=0.27T

Angle between the current and magnetic field, θ=90∘θ=90∘

(Since the magnetic field produced by a solenoid is along its axis and current carrying wire is kept perpendicular to the axis)

The magnetic force exerted on the wire is given as,

F=BIlsinθF=BIlsinθ

Substituting the given values,

F=0.27×10×0.03sin90∘F=0.27×10×0.03sin90∘

⇒F=8.1×10−2N⇒F=8.1×10−2N

Clearly, the magnetic force on the wire is found to be 8.1×10−2N8.1×10−2N. The direction of the force can be obtained from Fleming’s left-hand rule.

Current flowing in wire A, IA=8.0AIA=8.0A

Current flowing in wire B, IB=5.0AIB=5.0A

Distance between the two wires, r=4.0cm=0.04mr=4.0cm=0.04m

Length of a section of wire A, l=10cm=0.1ml=10cm=0.1m

Force exerted on length ll due to the magnetic field is given as,

B=2μ0IAIBl4πrB=2μ0IAIBl4πr

Where, μ0=4π×10−4TmA−1μ0=4π×10−4TmA−1is the permeability of free space.

On substituting the given values, we get,

B=4π×10−7×2×8×5×0.14π×0.04B=4π×10−7×2×8×5×0.14π×0.04

⇒B=2×10−5N⇒B=2×10−5N

The magnitude of force is 2×10−5N2×10−5N. This is an attractive force that is normal to A towards B because the direction of the currents in the wires is the same.

Length of the solenoid, l=80cm=0.8ml=80cm=0.8m

Since there are five layers of windings of 400 turns each on the solenoid.

Total number of turns on the solenoid would be, N=5×400=2000N=5×400=2000

Diameter of the solenoid, D=1.8cm=0.018mD=1.8cm=0.018m

Current carried by the solenoid, I=8.0AI=8.0A

We have the magnitude of the magnetic field inside the solenoid near its centre given by the relation,

B=μ0NIlB=μ0NIl

Where, μ0=4π×10−4TmA−1μ0=4π×10−4TmA−1is the permeability of free space.

On substituting the given values we get,

B=4π×10−7×2000×80.8B=4π×10−7×2000×80.8

⇒B=2.512×10−2T⇒B=2.512×10−2T

Clearly, the magnitude of the magnetic field inside the solenoid near its centre is found to be 2.512×10−2T2.512×10−2T.

Length of a side of the square coil, l=10cm=0.1ml=10cm=0.1m

Area of the square, A=l2=(0.1)2=0.01m2A=l2=(0.1)2=0.01m2

Current flowing in the coil, I=12AI=12A

Number of turns on the coil, n=20n=20

Angle made by the plane of the coil with magnetic field, θ=30∘θ=30∘

Strength of magnetic field, B=0.80TB=0.80T

Magnitude of the magnetic torque experienced by the coil in the magnetic field is given by the relation,

τ=nIABsinθτ=nIABsinθ

Substituting the given values, we get,

τ=20×0.8×12×0.01×sin30∘τ=20×0.8×12×0.01×sin30∘

⇒τ=0.96Nm⇒τ=0.96Nm

Thus, the magnitude of the torque experienced by the coil is 0.96 N m.

R1=10ΩR1=10Ω , N1=30N1=30, A1=3.6×10−3m2A1=3.6×10−3m2 , B1=0.25TB1=0.25T,R2=14ΩR2=14Ω ,N2=42N2=42A2=1.8×10−3m2A2=1.8×10−3m2 , B2=0.50TB2=0.50T

(The spring constants are identical for the meters).

Determine the Ratio of:

a) Current Sensitivity of M2 and M1M2 and M1

For moving coil meter M1M1,

Resistance, R1=10ΩR1=10Ω

Number of turns, N1=30N1=30

Area of cross-section, A1=3.6×10−3m2A1=3.6×10−3m2

Magnetic field strength, B1=0.25TB1=0.25T

Spring constant, K1=KK1=K

For moving coil meter M2M2:

Resistance, R2=14ΩR2=14Ω

Number of turns, N2=42N2=42

Area of cross-section, A2=1.8×10−3m2A2=1.8×10−3m2

Magnetic field strength, B2=0.50TB2=0.50T

Spring constant, K2=KK2=K

Current sensitivity of M1M1 is given as:

IS1=N1B1A1K1IS1=N1B1A1K1

And, current sensitivity of M2M2 is given as:

IS2=N2B2A2K2IS2=N2B2A2K2

On taking the ratio, we get,

⇒IS2IS1=N2B2A2K2N1B1A1K1⇒IS2IS1=N2B2A2K2N1B1A1K1

Substituting the values we get,

⇒IS2IS1=42×0.5×1.8×10−3×10×K14×30×0.25×3.6×10−3×K⇒IS2IS1=42×0.5×1.8×10−3×10×K14×30×0.25×3.6×10−3×K

⇒IS2IS1=1.4⇒IS2IS1=1.4

Therefore, the ratio of current sensitivity of M2 and M1M2 and M1 is 1.4.

VS2=N2B2A2K2R2VS2=N2B2A2K2R2

And, voltage sensitivity for M1M1is given as:

VS1=N1B1A1K1R1VS1=N1B1A1K1R1

On taking the ratio we get,

⇒VS2VS1=N2B2A2K1R1K2R2N1B1A1⇒VS2VS1=N2B2A2K1R1K2R2N1B1A1

Substituting the given values, we get,

⇒VS2VS1=42×0.5×1.8×10−3×10×KK×14×30×0.25×3.6×10−3=1⇒VS2VS1=42×0.5×1.8×10−3×10×KK×14×30×0.25×3.6×10−3=1

Thus, the ratio of voltage sensitivity of M2 and M1M2 and M1is 1.

Speed of the electron, V=4.8×106m/sV=4.8×106m/s

Charge on the electron, e=1.6×10−19Ce=1.6×10−19C

Mass of the electron, me=9.1×10−31kgme=9.1×10−31kg

Angle between the shot electron and magnetic field, θ=90∘θ=90∘

Magnetic force exerted on the electron in the magnetic field could be given as:

F=evBsinθF=evBsinθ

This force provides centripetal force to the moving electron and hence, the electron starts moving in a circular path of radius r.

Hence, centripetal force exerted on the electron would be,

FC=mv2rFC=mv2r

However, we know that in equilibrium, the centripetal force exerted on the electron is equal to the magnetic force i.e.,

FC=FFC=F

⇒mv2r=evBsinθ⇒mv2r=evBsinθ

⇒r=mvBesinθ⇒r=mvBesinθ

Substituting the given values we get,

⇒r=9.1×10−31×4.8×1066.5×10−4×1.6×10−19×sin90∘⇒r=9.1×10−31×4.8×1066.5×10−4×1.6×10−19×sin90∘

⇒r=4.2cm⇒r=4.2cm

Clearly, we found the radius of the circular orbit to be 4.2cm.

Magnetic field strength, B=6.5×10−4TB=6.5×10−4T

Charge of the electron, e=1.6×10−19Ce=1.6×10−19C

Mass of the electron, me=9.1×10−31kgme=9.1×10−31kg

Velocity of the electron, v=4.8×106m/sv=4.8×106m/s

Radius of the orbit, r=4.2cm=0.042mr=4.2cm=0.042m

Frequency of revolution of the electron νν

Angular frequency of the electron ω=2πθω=2πθ

Velocity of the electron is related to the angular frequency as:

v=rωv=rω

In the circular orbit, the magnetic force on the electron provides the centripetal force. Hence,

evB=mv2revB=mv2r

⇒eB=mr(rω)=mr(r2πν)⇒eB=mr(rω)=mr(r2πν)

⇒ν=6.5×10−4×1.6×10−192×3.14×9.1×10−31⇒ν=6.5×10−4×1.6×10−192×3.14×9.1×10−31∴ν=18.2×106Hz≈18MHz∴ν=18.2×106Hz≈18MHz

Thus, the frequency of the electron is around 18 MHz and is independent of the speed of the electron.

a) A Circular Coil of 30 Turns and Radius 8.0cm8.0cm Carrying a Current of 6.0A6.0A is Suspended Vertically in a Uniform Horizontal Magnetic Field of Magnitude 1.0T1.0T The Field Lines Make an Angle of 60∘60∘ with the Normal of the Coil. Calculate the Magnitude of the Counter Torque That Must Be Applied to Prevent the Coil from Turning.

Radius of the coil, r=8.0cm=0.08mr=8.0cm=0.08m

Area of the coil, A=πr2=π(0.08)2=0.0201m2A=πr2=π(0.08)2=0.0201m2

Current flowing in the coil is given to be, I=6.0AI=6.0A

Magnetic field strength, B=1TB=1T

Angle between the field lines and normal with the coil surface, θ=60∘θ=60∘

The coil will turn when it experiences a torque in the magnetic field. The counter torque applied to prevent the coil from turning is given by the relation, τ=nIABsinθτ=nIABsinθ

⇒τ=30×6×1×0.0201×sin60∘⇒τ=30×6×1×0.0201×sin60∘

⇒τ=3.133Nm⇒τ=3.133Nm

On the other hand, it is dependent on the area of the coil.

Thus, we could say that the answer would not change if the circular coil in the above case is replaced by a planar coil of some irregular shape that encloses the same area.

Radius of coil X, r1=16cm=0.16mr1=16cm=0.16m

Radius of coil Y, r2=10cm=0.1mr2=10cm=0.1m

Number of turns of on coil X,n1=20n1=20

Number of turns of on coil Y,n2=25n2=25

Current in coil X, I1=16AI1=16A

Current in coil Y,I2=18AI2=18A

Magnetic field due to coil X at their centre is given by the relation,

B1=μ0n1I12r1B1=μ0n1I12r1

Where, Permeability of free space, μ0=4π×10−4TmA−1μ0=4π×10−4TmA−1

B1=4π×10−7×20×162×0.16B1=4π×10−7×20×162×0.16

⇒B1=4π×10−4T⇒B1=4π×10−4T(towards East)

Magnetic field due to coil Y at their centre is given by the relation,

B2=μ0n2I22r2B2=μ0n2I22r2

⇒B2=4π×10−7×25×182×0.10⇒B2=4π×10−7×25×182×0.10

⇒B2=9π×10−4T⇒B2=9π×10−4T(towards West)

Clearly, net magnetic field could be obtained as:

B=B2−B1=9π×10−4−4π×10−4B=B2−B1=9π×10−4−4π×10−4

⇒B=1.57×10−3T⇒B=1.57×10−3T(towards West)

Magnetic field strength,B=100G=100×10−4TB=100G=100×10−4T

Number of turns per unit length,n=1000turns per mn=1000turns per m

Current flowing in the coil,I=15AI=15A

Permeability of free space, μ0=4π×10−7TmA−1μ0=4π×10−7TmA−1

Magnetic field is given the relation,

B=μ0nIB=μ0nI

⇒nI=Bμ0=100×10−44π×10−7⇒nI=Bμ0=100×10−44π×10−7

⇒nI≈8000A/m⇒nI≈8000A/m

If the length of the coil is taken as 50 cm, radius 4 cm, number of turns 400, and current 10 A, then these values are not unique for the given purpose. There is always a possibility of some adjustments with limits.

a) Show that this reduces to the familiar result for the field at the centre of the coil.

Radius of circular coil = R

Number of turns on the coil = N

Current in the coil = I

Magnetic field at a point on its axis at distance x is given by the relation,

Where,μ0=4π×10−4TmA−1μ0=4π×10−4TmA−1Permeability of free space

If the magnetic field at the centre of the coil is considered, then x=0x=0

B=μ0IR2N2(x2+R2)32B=μ0IR2N2(x2+R2)32

This is the familiar result for the magnetic field at the centre of the coil.

Radii of two parallel co-axial circular coils = R

Number of turns on each coil = N

Current in both coils = I

Distance between both the coils = R

Let us consider point Q at distance d from the centre.

Then, one coil is at a distance of R2+dR2+dfrom point Q.

Magnetic field at point Q could be given as:

B=μ0IR2N2(x2+R2)32B=μ0IR2N2(x2+R2)32

Also, the other coil is at a distance of R2+dR2+dfrom point Q.

Magnetic field due to this coil is given as:

B2=μ0NIR22[(R2−d)2+R2]32B2=μ0NIR22[(R2−d)2+R2]32

Now we have the total magnetic field as,

B=B1+B2B=B1+B2

⇒B=μ0IR22[{{(R2−d)2+R2}−32+{(R2+d)2+R2}−32×N]⇒B=μ0IR22[{{(R2−d)2+R2}−32+{(R2+d)2+R2}−32×N]

⇒B=μ0IR22[{{5R24+d2−Rd}−32+{5R24+d2+Rd}−32×N]⇒B=μ0IR22[{{5R24+d2−Rd}−32+{5R24+d2+Rd}−32×N]

⇒B=μ0IR22[{{1+4d25R2−4d5R}−32+{1+4d25R2+4d5R}−32×N]⇒B=μ0IR22[{{1+4d25R2−4d5R}−32+{1+4d25R2+4d5R}−32×N]

Now for d≪Rd≪R, we could neglect the factor d2R2d2R2, we get,

B≈μ0IR22×(5R24)−32[(1−4d5R)−32+(1+4d5R)−32]×NB≈μ0IR22×(5R24)−32[(1−4d5R)−32+(1+4d5R)−32]×N

⇒B≈μ0IR22×(5R24)−32[1−6d5R+1+6d5R]⇒B≈μ0IR22×(5R24)−32[1−6d5R+1+6d5R]

⇒B≈(45)32μ0INR=0.72(μ0INR)⇒B≈(45)32μ0INR=0.72(μ0INR)

Clearly, we proved that the field along the axis around the mid-point between the coils is uniform.

a) Outside the Toroid

Inner radius of the toroid, r1=25cm=0.25mr1=25cm=0.25m

Outer radius of the toroid, r2=26cm=0.26mr2=26cm=0.26m

Number of turns on the coil, N=3500N=3500

Current in the coil, I=11AI=11A

Magnetic field outside a toroid is zero. It is non-zero only inside the core of a toroid.

Where, Permeability of free space μ0=4π×10−7TmA−1μ0=4π×10−7TmA−1

ll is the length of toroid, given by

l=2π(r1+r22)=π(0.25+0.26)=0.51πl=2π(r1+r22)=π(0.25+0.26)=0.51π

⇒B=4π×10−7×3500×110.51π≈3.0×10−2T⇒B=4π×10−7×3500×110.51π≈3.0×10−2T

Thus, magnetic field inside the core of the toroid is 3.0×10−2T3.0×10−2T.

a) A Magnetic Field That Varies in Magnitude from Point to Point but Has a Constant Direction (east to West) Is Set up in a Chamber. A Charged Particle Enters the Chamber and Travels Undeflected Along a Straight Path With Constant Speed. What Can You Say About the Initial Velocity of the Particle?

This moving electron stays undeflected when the electric force acting on it is equal and opposite of the magnetic field.

The magnetic force would stay directed towards the South. Also, according to Fleming’s left-hand rule, the magnetic field must be applied in a vertically downward direction.

a) Is Transverse to Its Initial Velocity.

Magnetic field strength, B=0.15TB=0.15T

Charge on the electron, e=1.6×10−19Ce=1.6×10−19C

Mass of the electron, m=9.1×10−31kgm=9.1×10−31kg

Potential difference, V=2.0kV=2×103VV=2.0kV=2×103V

Now we have the kinetic energy of the electron given by,

K.E=eVK.E=eV

Substituting the given values we get,

eV=12mv2eV=12mv2

⇒v=2eVm−−−√⇒v=2eVm…………………….. (1)

Where, vvis the velocity of the electron

Since the magnetic force on the electron provides the required centripetal force of the electron, the electron traces a circular path of radius rr.

Now, the magnetic force on the electron is given by the relation,

F=BevF=Bev

Centripetal force,

FC=mv2rFC=mv2r

⇒Bev=mv2r⇒Bev=mv2r

⇒r=mvBe⇒r=mvBe………………….. (2)

From the equations (1) and (2), we get,

r=mBe[2eVm]12r=mBe[2eVm]12

Substituting the given values,

⇒r=9.1×10−310.15×1.6×10−19(2×1.6×10−19×2×1039.1×10−31)12⇒r=9.1×10−310.15×1.6×10−19(2×1.6×10−19×2×1039.1×10−31)12

⇒r=100.55×10−5⇒r=100.55×10−5

⇒r=1mm⇒r=1mm

Thus, we found that the electron has a circular trajectory of radius 1.0 mm normal to the magnetic field.

v1=vsinθv1=vsinθ

From equation (2), we can write the following expression:

r1=mv1Ber1=mv1Be

⇒r1=mvsinθBe⇒r1=mvsinθBe

⇒r1=9.1×10−310.15×1.6×10−19[2×1.6×10−19×2×1039×10−31]sin30∘⇒r1=9.1×10−310.15×1.6×10−19[2×1.6×10−19×2×1039×10−31]sin30∘

⇒r=0.5×10−3m=0.5mm⇒r=0.5×10−3m=0.5mm

Clearly, we found that the electron has a helical trajectory of radius0.5mm0.5mm, with axis of the solenoid along the magnetic field direction.

Magnetic field, B=0.75TB=0.75T

Accelerating voltage, V=15kV=15×103VV=15kV=15×103V

Electrostatic field, E=9×105Vm−1E=9×105Vm−1

Mass of the electron=m=m

Charge of the electron =e=e

Velocity of the electron =v=v

Kinetic energy of the electron =eV=eV

Thus,

12mv2=eV12mv2=eV

⇒em=v22V⇒em=v22V…………………….. (1)

Since the particle remains undeflected by electric and magnetic fields, we could infer that the electric field is balancing the magnetic field.

eE=evBeE=evB

⇒v=EB⇒v=EB……………………….. (2)

Now we could substitute equation (2) in equation (1) to get,

em=12(EB)2V=E22VB2em=12(EB)2V=E22VB2

⇒em=(9.0×105)22×15000×(0.75)2=4.8×107C/kg⇒em=(9.0×105)22×15000×(0.75)2=4.8×107C/kg

This value of specific charge (em)(em) is equal to the value of deuteron or deuterium ions. This is not a unique answer. Other possible answers are He++He++, Li+++Li+++

a) What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero?

Length of the rod, l=0.45ml=0.45m

Mass suspended by the wires, m=60g=60×10−3kgm=60g=60×10−3kg

Acceleration due to gravity, g=9.8ms−2g=9.8ms−2

Current in the rod flowing through the wire, I=5AI=5A

We could say that magnetic field (B) is equal and opposite to the weight of the wire i.e.,

BIl=mgBIl=mg

⇒B=mgIl=60×10−3×9.85×0.45⇒B=mgIl=60×10−3×9.85×0.45

⇒B=0.26T⇒B=0.26T

Clearly, a horizontal magnetic field of 0.26 T normal to the length of the conductor should be set up.

T=0.26×5×0.45+(60×10−3)×9.8T=0.26×5×0.45+(60×10−3)×9.8

⇒T=1.176N⇒T=1.176N

Current in both wires, I=300AI=300A

Distance between the wires, r=1.5cm=0.015mr=1.5cm=0.015m

Length of the two wires, l=70cm=0.7ml=70cm=0.7m

We know that, Force between the two wires is given by the relation,

F=μ0I22πrF=μ0I22πr

Where, Permeability of free spaceμ0=4π×10TmA−1μ0=4π×10TmA−1

As the direction of the current in the wires is found to be opposite, a repulsive force exists between them.

a) The Wire Intersects the Axis,

Magnetic field strength, B=1.5TB=1.5T

Radius of the cylindrical region, r=10cm=0.1mr=10cm=0.1m

Current in the wire passing through the cylindrical region, I=7AI=7A

If the wire intersects the axis, then the length of the wire is the diameter of the cylindrical region. Thus, l=2r=0.2ml=2r=0.2m

Angle between magnetic field and current, θ=90∘θ=90∘

We know that, Magnetic force acting on the wire is given by the relation,

F=BIlsinθF=BIlsinθ

⇒F=1.5×7×0.2×sin90∘⇒F=1.5×7×0.2×sin90∘

⇒F=2.1N⇒F=2.1N

Clearly, a force of 2.1 N acts on the wire in a vertically downward direction.

l1=lsinθl1=lsinθ

Angle between magnetic field and current, θ=45∘θ=45∘

Force on the wire,

F=BIl1sinθ=BIl=1.5×7×0.2F=BIl1sinθ=BIl=1.5×7×0.2

⇒F=2.1N⇒F=2.1N

Thus, a force of 2.1 N acts vertically downward on the wire. This is independent of angle θθ as lsinθlsinθ is fixed.

Let l2l2be the new length of the wire,

(l22)2=4(d+r)=4(10+6)=4×16(l22)2=4(d+r)=4(10+6)=4×16

⇒l2=8×2=16cm=0.16m⇒l2=8×2=16cm=0.16m

Magnetic force that is exerted on the wire is,

F2=BIl2=1.5×7×0.16F2=BIl2=1.5×7×0.16

⇒F=1.68N⇒F=1.68N

Clearly, a force of 1.68N1.68Nacts in a vertically downward direction on the wire.

Magnetic field strength, B=3000G=3000×10−4T=0.3TB=3000G=3000×10−4T=0.3T

Length of the rectangular loop, l=10cml=10cm

Width of the rectangular loop, b=5cmb=5cm

Area of the loop, A=l×b=(10×5)cm2=50×10−4m2A=l×b=(10×5)cm2=50×10−4m2

Current in the loop, I=12AI=12A

Now, we could take the anti-clockwise direction of the current as positive and vice-versa,

We have the expression for torque given as,

τ⃗=IA⃗ ×B⃗ τ→=IA→×B→

We could see from the given figure that A is normal to the y-z plane and B is directed along the z-axis. Substituting the given values,

τ=12×(50×10−4)i^×0.3k^τ=12×(50×10−4)i^×0.3k^

⇒τ=−1.8×10−2j^Nm⇒τ=−1.8×10−2j^Nm

Now, the torque is found to be directed along negative y-direction. Since the external magnetic field is uniform, the force on the loop would be zero.

τ=IA⃗ ×B⃗ τ=IA→×B→

⇒τ=−12(50×10−4)j^×0.3k^⇒τ=−12(50×10−4)j^×0.3k^

⇒τ=−1.8×10−2i^Nm⇒τ=−1.8×10−2i^Nm

The direction here is along the negative x direction and the force is zero.

|τ|=IAB|τ|=IAB

⇒τ=12×(50×10−4)×0.3⇒τ=12×(50×10−4)×0.3

⇒|τ|=1.8×10−2Nm⇒|τ|=1.8×10−2Nm

Here, the direction is found to be at 240∘240∘with positive x-direction and the force is zero.

τ=IA⃗ ×B⃗ τ=IA→×B→

⇒τ=(50×10−4×12)k^×0.3k^⇒τ=(50×10−4×12)k^×0.3k^

⇒τ=0⇒τ=0

Here, both torque and force are found to be zero.

τ=IA⃗ ×B⃗ τ=IA→×B→

⇒τ=(50×10−4×12)k^×0.3k^⇒τ=(50×10−4×12)k^×0.3k^

⇒τ=0⇒τ=0

Here also both torque and force are found to be zero.

For the case (e) the direction IA⃗ IA→and B⃗ B→ is the same and the angle between them is zero. They would come back to equilibrium on being displaced and so the equilibrium is stable.

For the case (f), the direction of IA⃗ IA→and B⃗ B→ are opposite and the angle between them is 180∘180∘. Here it doesn’t come back to its original position on being disturbed and hence, the equilibrium is unstable.

a) Total Torque on the Coil?

Number of turns on the circular coil, n=20n=20

Radius of the coil, r=10cm=0.1mr=10cm=0.1m

Magnetic field strength, B=0.10TB=0.10T

Current in the coil, I=5.0AI=5.0A

As the angle between force and the normal to the loop is zero, the total torque on the coil is zero.

So, τ=NIABsinθτ=NIABsinθis zero.

Cross-sectional area of copper coil, A=10−5m2A=10−5m2

Number of free electrons per cubic meter in copper, N=1029/m3N=1029/m3

Charge on the electron would be, e=1.6×10−19Ce=1.6×10−19C

Magnetic force, F=BevdF=Bevd

Where, vdvd is drift velocity of electrons given by INeAINeA

⇒F=BeINeA=0.10×5.01029×10−5=5×10−25N⇒F=BeINeA=0.10×5.01029×10−5=5×10−25N

Clearly, the average force on each electron is 5×10−25N5×10−25N.

Length of the solenoid, L=60cm=0.6mL=60cm=0.6m

Radius of the solenoid, r=4.0cm=0.04mr=4.0cm=0.04m

It is given that there are 3 layers of windings of 300 turns each.

Total number of turns, n=3×300=900n=3×300=900

Length of the wire, l=2cm=0.02ml=2cm=0.02m

Mass of the wire, m=2.5g=2.5×10−3kgm=2.5g=2.5×10−3kg

Current flowing through the wire, i=6Ai=6A

Acceleration due to gravity, g=9.8ms−2g=9.8ms−2

Magnetic field produced inside the solenoid, B=μ0nILB=μ0nIL

Where, Permeability of free space μ0=4π×10−7TmA−1μ0=4π×10−7TmA−1

Current flowing through the windings of the solenoid, II

Magnetic force is given by the relation,

F=Bil=μ0nILilF=Bil=μ0nILil

Now, we have the force on the wire equal to the weight of the wire.

mg=μ0nIilLmg=μ0nIilL

⇒I=mgLμ0nil=2.5×10−3×9.8×0.64π×10−7×900×0.02×6⇒I=mgLμ0nil=2.5×10−3×9.8×0.64π×10−7×900×0.02×6

⇒I=108A⇒I=108A

Clearly, the current flowing through the solenoid is 108 A.

Resistance of the galvanometer coil, G=12ΩG=12Ω

Current for which there is full scale deflection, Ig=3mA=3×10−3AIg=3mA=3×10−3A

Range of the voltmeter needs to be converted to 18V18V.

Let a resistor of resistance R be connected in series with the galvanometer to convert it into a voltmeter. This resistance can be given as:

R=VIg−GR=VIg−G

Substituting the given values we get,

R=183×10−3−12=6000−12R=183×10−3−12=6000−12

⇒R=5988Ω⇒R=5988Ω

Clearly, we found that a resistor of resistance 5988Ω5988Ωis to be connected in series with the given galvanometer.

Resistance of the galvanometer coil, G=15ΩG=15Ω

Current for which the galvanometer shows full scale deflection, Ig=4mA=4×10−3AIg=4mA=4×10−3A

We said that, Range of the ammeter needs to be 6A6A.

In order to convert the given galvanometer into an ammeter, a shunt resistor of resistance S is to be connected in parallel with the galvanometer.

The value of S could be given as:

S=IgGI−IgS=IgGI−Ig

Substituting the given values we get,

S=4×10−3×156−4×10−3=0.065.996≈0.01ΩS=4×10−3×156−4×10−3=0.065.996≈0.01Ω

⇒S=10mΩ⇒S=10mΩ

Clearly, we found that a

10mΩ10mΩ

shunt resistor is to be connected in parallel with the galvanometer.

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