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

1. A Circular Coil of Wire Consisting of 100100 Turns, Each of Radius 8.0cm8.0cm Carries a Current of 0.40A0.40A. What is the Magnitude of the Magnetic Field B at the Centre of the Coil?

Ans: We are given:
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.

2. A Long Straight Wire Carries a Current of 35A35A. What Is the Magnitude of Field B at a Point 20cm from the Wire?

Ans: We are given the following:
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.

3. A Long Straight Wire in the Horizontal Plane Carries a Current of 50A50A in North to South Direction. Give the Magnitude and Direction of B at a Point 2.5m2.5m East of the Wire.

Ans: We are given the following:
The current in the wire, I=50AI=50A
The distance of the given point from the wire, r=2.5mr=2.5m
(Image will be Uploaded Soon)
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.

4. A Horizontal Overhead Power Line Carries a Current of 90A90A in East to West Direction. What is the Magnitude and Direction of the Magnetic Field Due to the Current 1.5m1.5m Below the Line?

Ans: We are given the following:
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.

5. What is the Magnitude of Magnetic Force Per Unit Length on a Wire Carrying a Current of 8 and Making an Angle of 30∘30∘ with the Direction of a Uniform Magnetic Field of 0.15T0.15T?

Ans: Given that,
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=BIsin⁡θ
Substituting the given values, we get,
F=0.15×8×1×sin30∘F=0.15×8×1×sin⁡30∘
⇒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

6. A 3.0cm3.0cm Wire Carrying a Current of 10A10A is Placed Inside a Solenoid Perpendicular to Its Axis. The Magnetic Field Inside the Solenoid Is Given to Be 0.27T0.27T. What is the Magnetic Force on the Wire?

Ans: We are given the following,
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=BIlsin⁡θ
Substituting the given values,
F=0.27×10×0.03sin90∘F=0.27×10×0.03sin⁡90∘
⇒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.

7. Two Long and Parallel Straight Wires A and B Carrying Currents of 8.0A8.0Aand 5.0A5.0A in the Same Direction are Separated by a Distance of 4.0cm4.0cm. Estimate the Force on a 10cm10cm Section of Wire A.

Ans: We are given:
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.

8. A Closely Wound Solenoid 80cm80cm Long has 55 Layers of Windings of 400400 Turns Each. The Diameter of the Solenoid is 1.8cm1.8cm. If the Current Carried is 8.0A8.0A, Estimate the Magnitude of B Inside the Solenoid Near its Centre.

Ans: We are given the following:
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.

9. A Square Coil of Side 10cm10cm Consists of 20 Turns and Carries a Current of 12A12A. The Coil Is Suspended Vertically and the Normal to the Plane of the Coil Makes an Angle of 30∘30∘ with the Direction of a Uniform Horizontal Magnetic Field of Magnitude 0.80T0.80T. What is the Magnitude of Torque Experienced by the Coil?

Ans: We are given the following:
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=nIABsin⁡θ
Substituting the given values, we get,
τ=20×0.8×12×0.01×sin30∘τ=20×0.8×12×0.01×sin⁡30∘
⇒τ=0.96Nm⇒τ=0.96Nm
Thus, the magnitude of the torque experienced by the coil is 0.96 N m.

10. Two Moving Coil Meters, M1M1 and M2M2 Have the Following Particulars:
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

Ans: We are given:
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.

b) Voltage Sensitivity of M2 and M1M2 and M1

Ans: Voltage sensitivity for M2M2is given is:
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.

11. In a Chamber, a Uniform Magnetic Field of 6.5G(1G=10−4T)6.5G(1G=10−4T)is Maintained. An Electron Is Shot Into the Field With a Speed Of 4.8×106ms−14.8×106ms−1 Normal to the Field. Explain Why the Path of the Electron is a Circle. Determine the Radius of the Circular Orbit.(e=1.6×10−19C,me=9.1×10−31kg)(e=1.6×10−19C,me=9.1×10−31kg)

Ans: Magnetic field strength, B=6.5G=6.5×10−4TB=6.5G=6.5×10−4T
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=evBsin⁡θ
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=evBsin⁡θ
⇒r=mvBesinθ⇒r=mvBesin⁡θ
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×sin⁡90∘
⇒r=4.2cm⇒r=4.2cm
Clearly, we found the radius of the circular orbit to be 4.2cm.

12. In Exercise 4.11 Obtain the Frequency of Revolution of the Electron in Its Circular Orbit. Does the Answer Depend on the Speed of the Electron? Explain.

Ans: We are given the following:
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.

13.
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.

Ans: Number of turns on the circular coil, n=30n=30
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=nIABsin⁡θ
⇒τ=30×6×1×0.0201×sin60∘⇒τ=30×6×1×0.0201×sin⁡60∘
⇒τ=3.133Nm⇒τ=3.133Nm

b) Would Your Answer Change, If the Circular Coil in (a) Were Replaced by a Planar Coil of Some Irregular Shape that Encloses the Same Area? (all Other Particulars Are Also Unaltered).

Ans: From the part(a) we could infer that the magnitude of the applied torque is not dependent on the shape of the coil.
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.

14. Two Concentric Circular Coils X and Y Radii 16 cm and 10 cm, Respectively, Lie in the Same Vertical Plane Containing the North to South Direction. Coil X has 20 Turns and Carries a Current of 16 A; Coil Y has 25 Turns and Carries a Current Of 18 A. The Sense of the Current in X is Anticlockwise, and Clockwise in Y, for an Observer Looking at the Coils Facing West. Give the Magnitude and Direction of the Net Magnetic Field Due to the Coils at Their Centre.

Ans: We are given,
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)

15. A magnetic field of 100G100G(where, 1G=10−4T)(where, 1G=10−4T) is required which is uniform in a region of linear dimension about 10cm10cm and area of cross-section about10−3m210−3m2. The maximum current carrying capacity of a given coil of wire is 15A15A and the number of turns per unit length that can be wound a core is at most 1000 turns per m1000 turns per m. Suggest some appropriate design particulars to a solenoid for the required purpose. Assume the core is not ferromagnetic.

Ans: We are given,
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.

16. For a Circular Coil of Radius R and N Turns Carrying Current I, the Magnitude of the Magnetic Field at a Point on Its Axis at a Distance X from Its Centre Is Given By,
a) Show that this reduces to the familiar result for the field at the centre of the coil.

Ans: We are given,
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.

b) Consider Two Parallel Co-axial Circular Coils of Equal Radius R, and the Number of Turns N, Carrying Equal Currents in the Same Direction, and Separated by a Distance R. Show That the Field on the Axis Around the Mid-point Between the Coils Is Uniform Over a Distance That Is Small as Compared to R, and Is Given By, Approximately, (such an Arrangement to Produce a Nearly Uniform Magnetic Field Over a Small Region Is Known as Helmholtz Coils.)

Ans: Given that,
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.

17. A Toroid Has a Core (non-Ferromagnetic) of Inner Radius 25 Cm and Outer Radius 26 Cm, Around Which 3500 Turns of a Wire Are Wound. If the Current in the Wire is 11 A, What is the Magnetic Field
a) Outside the Toroid

Ans: We are given,
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.

b) Inside the Core of the Toroid.

Ans: Magnetic field inside the core of a toroid is given by the relation, B=μ0NIlB=μ0NIl
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.

c) In the Empty Space Surrounded by the Toroid.

Ans: The empty space that is surrounding the toroid has a magnetic field equal to zero.

18. Answer the Following Questions:
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?

Ans: The initial velocity of the particle could either be parallel or be anti-parallel to the magnetic field. So, it travels along a straight path without undergoing any deflection in the field.

b) A Charged Particle Enters an Environment of a Strong and Non-Uniform Magnetic Field Varying from Point to Point Both in Magnitude and Direction, and Comes Out of it Following a Complicated Trajectory. Would Its Final Speed Equal the Initial Speed If it Suffered No Collisions With the Environment?

Ans: Yes, the final speed of the charged particle would be equal to its initial speed as the magnetic force has the potential to change the direction of velocity even though not its magnitude.

c) An Electron Travelling West to East Enters a Chamber Having a Uniform Electrostatic Field in the North to South Direction. Specify the Direction in Which a Uniform Magnetic Field Should Be Set up to Prevent the Electron from Deflecting from Its Straight-Line Path.

Ans: An electron travelling from West to East enters a chamber having a uniform electrostatic field along the North-South direction.
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.

19. An Electron Emitted by a Heated Cathode and Accelerated Through a Potential Difference of 2.0kV2.0kV, Enters a Region With Uniform Magnetic Field of 0.15T0.15T. Determine the Trajectory of the Electron If the Field
a) Is Transverse to Its Initial Velocity.

Ans: We are given,
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.

b) Makes an Angle of 30∘30∘ with the Initial Velocity.

Ans: When the field makes an angle θθ of 30∘30∘with initial velocity, the initial velocity will be,
v1=vsinθv1=vsin⁡θ
From equation (2), we can write the following expression:
r1=mv1Ber1=mv1Be
⇒r1=mvsinθBe⇒r1=mvsin⁡θBe
⇒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]sin⁡30∘
⇒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.

20. A Magnetic Field Set up Using Helmholtz Coils (Described in Exercise 4.164.16) is Uniform in a Small Region and Has a Magnitude of 0.75T0.75T. In the Same Region, a Uniform Electrostatic Field Is Maintained in a Direction Normal to the Common Axis of the Coils. A Narrow Beam of (single Species) Charged Particles All Accelerated Through 15kV15kV Enters This Region in a Direction Perpendicular to Both the Axis of the Coils and the Electrostatic Field. If the Beam Remains Undeflected When the Electrostatic Field Is 9.0×10−5Vm−19.0×10−5Vm−1 Make a Simple Guess as to What the Beam Contains. Why Is the Answer Not Unique?

Ans: We are given,
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+++

21. A Straight Horizontal Conducting Rod of Length 0.45m0.45m and Mass 60g60g is Suspended by Two Vertical Wires at Its Ends. A Current of 5.0A5.0A is Set up in the Rod Through the Wires.
a) What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero?

Ans: We are given,
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.

b) What will be the Total Tension in the Wires If the Direction of Current is Reversed Keeping the Magnetic Field Same as Before? (Ignore the Mass of the Wires.) g=9.8ms−2g=9.8ms−2

Ans: When the direction of the current is reversed, BIlBIl and mgmg will act downwards. Clearly, the effective tension in the wires is found to be,
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

22. The Wires Which Connect the Battery of an Automobile to Its Starting Motor Carry a Current of 300A300A(for a short time). What Is the Force Per Unit Length Between the Wires if they are 70cm70cm Long and 1.5cm1.5cm Apart? Is the Force Attractive or Repulsive?

Ans: We are given,
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.

23. A Uniform Magnetic Field of 1.5T1.5T Exists in a Cylindrical Region of Radius 10.0cm10.0cm, its Direction Parallel to the Axis Along East to West. A Wire Carrying Current of 7.0A7.0A in the North to South Direction Passes Through This Region. What is the Magnitude and Direction of the Force on the Wire If,
a) The Wire Intersects the Axis,

Ans: We are given,
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=BIlsin⁡θ
⇒F=1.5×7×0.2×sin90∘⇒F=1.5×7×0.2×sin⁡90∘
⇒F=2.1N⇒F=2.1N
Clearly, a force of 2.1 N acts on the wire in a vertically downward direction.

b) The Wire is Turned from N-S to northeast-northwest Direction,

Ans: The new length of the wire after turning it to the northeast-northwest direction can be given as:
l1=lsinθl1=lsin⁡θ
Angle between magnetic field and current, θ=45∘θ=45∘
Force on the wire,
F=BIl1sinθ=BIl=1.5×7×0.2F=BIl1sin⁡θ=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 lsinθlsin⁡θ is fixed.

c) The Wire in the N-S Direction is Lowered from the Axis by a Distance of 6.0cm6.0cm?

Ans: The wire is lowered from the axis by distance, d=6.0cmd=6.0cm
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.

24. A Uniform Magnetic Field of 3000G3000G is Established Along the Positive Z-Direction. A Rectangular Loop of Sides 10cm10cm and 5cm5cm Carries a Current of 12A12A. What is the Torque on the Loop in the Different Cases Shown in Figure? What Is the Force on Each Case? Which Case Corresponds to Stable Equilibrium?

a)

Ans: We are given,
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.

b)

Ans: This case is very similar to case (a), and hence, the answer here would be same as (a).

c)

Ans: Torque here would be,
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.

d)

Ans: Torque here would be,
|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.

e)

Ans: Torque,
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.

f)

Ans: Torque is given by,
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.

25. A Circular Coil of 20 Turns and Radius 10 Cm is Placed in a Uniform Magnetic Field of 0.10 T Normal to the Plane of the Coil. If the Current in the Coil Is 5.0 A, What is the: (The Coil is Made of Copper Wire of Cross-Sectional Area 10−5m210−5m2, and the Free Electron Density in Copper Is Given to be about1029m−31029m−3).
a) Total Torque on the Coil?

Ans: We are given,
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=NIABsin⁡θis zero.

b) Total Force on the Coil,

Ans: There is no total force on the coil because the field is uniform.

C) Average Force on Each Electron in the Coil Due to the Magnetic Field?

Ans: Given that,
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.

26. A Solenoid 60cm60cm Long and Radius 4.0cm4.0cmhas 3 Layers of Windings of 300turns Each. A 2.0cm2.0cm Long Wire of Mass 2.5g2.5g Lies Inside the Solenoid (near Its Centre) Normal to Its Axis; Both the Wire and the Axis of the Solenoid Are in the Horizontal Plane. the Wire Is Connected Through Two Leads Parallel to the Axis of the Solenoid to an External Battery Which Supplies a Current of 6.0a6.0a in the Wire. What Value of Current (with Appropriate Sense of Circulation) in the Windings of the Solenoid Can Support the Weight of the Wire? g=9.8ms−2g=9.8ms−2

Ans: We are given:
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.

27. A Galvanometer Coil Has a Resistance of 12Ω12Ωand the Metre Shows Full Scale Deflection for a Current of 3mA3mA. How will you Convert the Metre Into a Voltmeter of Range 00 to 18V18V?

Ans: We are given,
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.

28. A Galvanometer Coil Has a Resistance of 15Ω15Ω and the Metre Shows Full Scale Deflection for a Current of 4mA4mA. How will you Convert the Metre Into an Ammeter of Range 00 to 6A6A?

Ans: We are given,
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.