Question:

A galvanometer coil has a resistance of 15 Ω and the metre shows full scale deflection for a current of 4 mA. How will you convert the metre into an ammeter of range 0 to 6 A?

Answer:

1. Calculate the current sensitivity of the galvanometer coil by dividing the full scale deflection current (4 mA) by the resistance of the coil (15 Ω): Current sensitivity = 4 mA / 15 Ω = 0.267 mA/Ω

2. Calculate the required shunt resistance for the ammeter by dividing the desired full scale deflection current (6 A) by the current sensitivity of the galvanometer coil (0.267 mA/Ω): Required shunt resistance = 6 A / 0.267 mA/Ω = 22,400 Ω

3. Connect the shunt resistance in series with the galvanometer coil to convert it into an ammeter of range 0 to 6 A.

Question:

What was Vijay Singh’s weakness? Which awkward situation did it push him into?

Answer:

Step 1: Research Vijay Singh’s background.

Step 2: Look for any information regarding his weaknesses.

Step 3: Read any available accounts of awkward situations he has been in.

Question:

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, B=μ0​IR^2N/2(x^2+R^2)^3/2​ (a) Show that this reduces to the familiar result for field at the centre of the coil. (b) Consider two parallel co-axial circular coils of equal radius R,and 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. B=0.72μ0​NI​/R [Such an arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.]

Answer:

(a) At the centre of the coil, x=0, so B = μ0​IR^2N/2(0+R^2)^3/2 = μ0​IN/2R

(b) At the mid-point of the two coils, x=0, so B = μ0​IR^2N/2(0+R^2)^3/2 = μ0​IN/2R

Since the two coils have the same current and number of turns, B = 0.72μ0​NI​/R

Question:

A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30o with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?

Answer:

Answer: Step 1: Calculate the area of the coil. Area of the coil = (Side)2 = (10 cm)2 = 100 cm2

Step 2: Calculate the magnetic flux. Magnetic flux = B × Area = 0.80 T × 100 cm2 = 80 mWb

Step 3: Calculate the magnetic moment. Magnetic moment = N × I × Area = 20 × 12 A × 100 cm2 = 2400 A m2

Step 4: Calculate the torque. Torque = Magnetic moment × Sin(30o) = 2400 A m2 × Sin(30o) = 1200 A m2

Question:

Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.

Answer:

1. Calculate the magnitude of the magnetic field produced by wire A using the equation B = μ0I/2πr, where μ0 is the permeability of free space (4π10^-7 Tm/A), I is the current in wire A (8.0 A), and r is the distance between wires A and B (4.0 cm).

2. Calculate the magnitude of the force on a 10 cm section of wire A using the equation F = BIL, where B is the magnetic field calculated in step 1, I is the current in wire A (8.0 A), and L is the length of the section of wire A (10 cm).

Question:

An electron emitted by a heated cathode and accelerated through a potential difference of 2.0 kV, enters a region with uniform magnetic field of 0.15 T. Determine the trajectory of the electron if the field (a) is transverse to its initial velocity, (b) makes an angle of 30∘ with the initial velocity.

Answer:

a) If the magnetic field is transverse to the initial velocity of the electron, the trajectory of the electron will form a circular path.

b) If the magnetic field makes an angle of 30° with the initial velocity of the electron, the trajectory of the electron will form a helical path.

Question:

A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?

Answer:

Answer:

1. Calculate the magnetic field strength: The magnetic field strength due to a current-carrying wire is given by the equation B = μoI/2πr, where μo is the permeability of free space (4π10^-7 Tm/A), I is the current in the wire (90 A), and r is the distance from the wire (1.5 m).

Therefore, the magnetic field strength is B = 4π*10^-7 * 90 / (2π * 1.5) = 0.6 T.

2. Calculate the direction of the magnetic field: The direction of the magnetic field due to a current-carrying wire is given by the right-hand rule. This states that if the thumb of the right hand is pointed in the direction of the current, then the fingers curl in the direction of the magnetic field.

Therefore, the direction of the magnetic field due to the current is from south to north.

Question:

The wires which connect the battery of an automobile to its starting motor carry a current of 300 A (for a short time). What is the force per unit length between the wires if they are 70 cm long and 1.5 cm apart? Is the force attractive or repulsive?

Answer:

Step 1: Calculate the force between the wires using Coulomb’s Law: F = k(Q1Q2/d^2), where k is the Coulomb’s constant (8.99 x 10^9 Nm^2/C^2), Q1 and Q2 are the charges of the wires (300A x 70 cm = 21000 C), and d is the distance between the wires (1.5 cm).

Step 2: Calculate the force: F = 8.99 x 10^9 Nm^2/C^2 x (21000 C)^2 / (1.5 cm)^2 = 9.3 x 10^11 N/m.

Step 3: Determine if the force is attractive or repulsive. Since the charges of the wires are both positive, the force between them is attractive.

Question:

A galvanometer coil has a resistance of 12 Ω and the metre shows full scale deflection for a current of 3 mA. How will you convert the metre into a voltmeter of range 0 to 18 V?

Answer:

1. Calculate the full-scale deflection (FSD) voltage of the galvanometer by multiplying the resistance of the coil (12 Ω) by the current (3 mA) to get 36 mV.

2. Calculate the resistance of the shunt required to convert the galvanometer into a voltmeter of range 0 to 18 V. This can be done by dividing the FSD voltage (36 mV) by the desired maximum voltage (18 V) to get a shunt resistance of 2 Ω.

3. Connect the shunt in parallel with the galvanometer coil.

4. The galvanometer is now converted into a voltmeter with a range of 0 to 18 V.

Question:

A long straight wire carries a current of 35 A. What is the magnitude of the field B at a point 20 cm from the wire?

Answer:

1. Calculate the permeability of free space, μ0, which is equal to 4π x 10-7 Tm/A.

2. Calculate the magnitude of the field B, which is equal to μ0 x I/2πr, where I is the current in the wire (35 A) and r is the distance from the wire (20 cm).

3. Plug the values into the equation and solve for B. B = 4π x 10-7 Tm/A x 35 A/ (2π x 0.2 m) = 0.875 T.

Question:

A magnetic field of 100 G (1 G = 10^−4 T) is required which is uniform in a region of linear dimension about 10 cm and area of cross-section about 10^−3m^2. The maximum current-carrying capacity of a given coil of wire is 15 A and the number of turns per unit length that can be wound round a core is at most 1000 turns m^−1. Suggest some appropriate design particulars of a solenoid for the required purpose. Assume the core is not ferromagnetic.

Answer:

1. Calculate the number of turns required for the solenoid: Number of turns = (Magnetic Field Strength x Area of Cross-Section)/(Turns per Unit Length x Current-Carrying Capacity) = (100 G x 10^-3 m^2)/(1000 turns m^-1 x 15 A) = 0.7 turns

2. Calculate the length of the solenoid: Length = Number of Turns/Turns per Unit Length = 0.7 turns/1000 turns m^-1 = 0.7 m

3. Calculate the diameter of the solenoid: Diameter = 2 x Linear Dimension = 2 x 10 cm = 20 cm

4. Calculate the required current: Current = Magnetic Field Strength x Area of Cross-Section/Number of Turns = (100 G x 10^-3 m^2)/0.7 turns = 14.3 A

Question:

A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire?

Answer:

1. Calculate the magnetic flux, Φ, through the solenoid using the equation Φ = BA, where B is the magnetic field inside the solenoid and A is the area of the solenoid:

Φ = (0.27 T)(π(1.5 cm)^2) = 2.37 x 10^-3 T m^2

2. Calculate the magnetic force, F, using the equation F = IΦL, where I is the current through the wire, Φ is the magnetic flux, and L is the length of the wire:

F = (10 A)(2.37 x 10^-3 T m^2)(3.0 cm) = 7.11 x 10^-3 N

Question:

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? (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? (c) An electron travelling west to east enters a chamber having a uniform electrostatic field in north to south direction. Specify the direction in which a uniform magnetic field should be setup to prevent the electron from deflecting from its straight line path.

Answer:

a) The initial velocity of the particle must be in the same direction as the constant direction of the magnetic field (east to west).

b) No, its final speed would not equal its initial speed if it suffered no collisions with the environment.

c) A uniform magnetic field should be set up in the east to west direction to prevent the electron from deflecting from its straight line path.

Question:

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, (b) inside the core of the toroid, and (c) in the empty space surrounded by the toroid.

Answer:

Step 1: Calculate the total magnetic flux of the toroid.

Magnetic flux = (N * I) * (2 * pi * r)

= (3500 * 11) * (2 * pi * 25)

= 6,746,000 mWb

Step 2: Calculate the magnetic field outside the toroid.

Magnetic field = (magnetic flux) / (2 * pi * r)

= (6,746,000) / (2 * pi * 26)

= 1,085 A/m

Step 3: Calculate the magnetic field inside the core of the toroid.

Magnetic field = (magnetic flux) / (2 * pi * r)

= (6,746,000) / (2 * pi * 25)

= 1,136 A/m

Step 4: Calculate the magnetic field in the empty space surrounded by the toroid.

Magnetic field = 0 A/m

Question:

In a chamber, a uniform magnetic field of 6.5 G (1G=10^−4T) is maintained. An electron is shot into the field with a speed of 4.8×106ms^−1 normal to the field. Explain why the path of the electron is a circle? Determine the radius of the circular orbit. (e=1.5×10^−19C,me​=9.1×10^−31kg). Also obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.

Answer:

1. The path of the electron is a circle because when an electron is placed in a uniform magnetic field, it experiences a force (known as the Lorentz force) which is always perpendicular to its velocity. This force causes it to move in a circular path.

2. The radius of the circular orbit can be determined using the equation F = (eBv)/(2πme), where F is the force, e is the charge of the electron, B is the magnetic field, v is the velocity of the electron, and me is the mass of the electron. Substituting the given values, we get F = (1.5 × 10^-19C × 6.5 G × 4.8 × 10^6ms^-1)/(2π × 9.1 × 10^-31kg) = 2.3 × 10^-15N. The radius of the circular orbit can then be calculated as r = (2πmev)/(eB) = (2π × 9.1 × 10^-31kg × 4.8 × 10^6ms^-1)/(1.5 × 10^-19C × 6.5 G) = 1.2 × 10^-5m.

3. The frequency of revolution of the electron in its circular orbit can be determined using the equation f = (eB)/(2πme), where f is the frequency, e is the charge of the electron, B is the magnetic field, and me is the mass of the electron. Substituting the given values, we get f = (1.5 × 10^-19C × 6.5 G)/(2π × 9.1 × 10^-31kg) = 3.2 × 10^7s^-1.

4. The answer does not depend on the speed of the electron, as the frequency of revolution is determined solely by the charge and mass of the electron and the strength of the magnetic field.

Question:

A long straight wire in the horizontal plane carries a current of 50 A in north to south direction. Give the magnitude and direction of B at a point 2.5 m east of the wire.

Answer:

1. The magnitude of the magnetic field at the point 2.5 m east of the wire is given by B = μoI/2πr, where μo is the permeability of free space (4π10^-7 Tm/A), I is the current in the wire (50 A) and r is the distance from the wire (2.5 m).

2. Therefore, the magnitude of the magnetic field at this point is B = 4π*10^-7 Tm/A * 50 A / 2π * 2.5 m = 0.0000625 T.

3. The direction of the magnetic field at this point is perpendicular to the plane of the wire and is given by the right-hand rule. This means that the direction of the magnetic field is pointing out of the plane of the wire, which is east.

Question:

What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of 30o with the direction of a uniform magnetic field of 0.15 T?

Answer:

1. Calculate the magnitude of the magnetic force: F = B I L sinθ

F = (0.15 T) (8 A) (1 m) (sin 30o)

F = 0.6 N

Question:

Two moving coil meters, M1​ and M2​ have the following particulars: R1​=10Ω,N1​=30 A1​=3.6×10−3m2,B1​=0.25T R2​=14Ω,N2​=42 A2​=1.8×10−3,B2​=0.50T (The spring constants are identical for the two meters).Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M2​ and M1​.

Answer:

(a) Current Sensitivity:

The current sensitivity of a moving coil meter is given by the formula: Current Sensitivity = (N * B) / (R * A)

Therefore, the current sensitivity of M1 is: Current Sensitivity of M1 = (30 * 0.25) / (10 * 3.6 x 10-3) = 0.694

The current sensitivity of M2 is: Current Sensitivity of M2 = (42 * 0.50) / (14 * 1.8 x 10-3) = 1.5

The ratio of current sensitivity of M2 and M1 is: Current Sensitivity Ratio = 1.5 / 0.694 = 2.16

(b) Voltage Sensitivity:

The voltage sensitivity of a moving coil meter is given by the formula: Voltage Sensitivity = (N * B) / R

Therefore, the voltage sensitivity of M1 is: Voltage Sensitivity of M1 = (30 * 0.25) / 10 = 0.075

The voltage sensitivity of M2 is: Voltage Sensitivity of M2 = (42 * 0.50) / 14 = 0.179

The ratio of voltage sensitivity of M2 and M1 is: Voltage Sensitivity Ratio = 0.179 / 0.075 = 2.38

Question:

A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?

Answer:

1. Calculate the area of the coil: A = πr2 = π(8.0 cm)2 = 201.06 cm2

2. Calculate the magnitude of the magnetic field: B = μoI/2πr = (4π10-7 Tm/A)(0.40 A)/(2π8 cm) = 0.0050 T

Question:

In a chamber, a uniform magnetic field of 6.5 G (1G=10^−4T) is maintained. An electron is shot into the field with a speed of 4.8×106ms−1 normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. (e=1.5×10−19C,me​=9.1×10^−31kg)

Answer:

Step 1: Magnetic force is given by F = qvB, where q is the charge of the electron, v is the velocity of the electron, and B is the magnetic field.

Step 2: Since the magnetic field is uniform, the force on the electron is constant, causing it to move in a circular path.

Step 3: The radius of the circular orbit can be determined by using the equation F = mv2/r, where m is the mass of the electron and r is the radius of the circular orbit.

Step 4: Substituting the given values into the equation, we get:

F = qvB = (1.5 × 10−19C)(4.8 × 106ms−1)(6.5 × 10−4T) = 4.68 × 10−13 N

Step 5: Substituting the given values into the equation F = mv2/r, we get:

4.68 × 10−13 N = (9.1 × 10−31kg)(4.8 × 106ms−1)2/r

Step 6: Solving for r, we get:

r = (9.1 × 10−31kg)(4.8 × 106ms−1)2/4.68 × 10−13 N = 3.2 × 10−12m

Therefore, the radius of the circular orbit is 3.2 × 10−12m.

Question:

A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of B inside the solenoid near its centre.

Answer:

1. Calculate the number of turns in the solenoid: 5 layers of 400 turns each = 2000 turns

2. Calculate the mean radius of the solenoid: (1.8 cm + 80 cm)/2 = 40.9 cm

3. Calculate the magnetic field using the equation B = μ₀NI/2πr: B = (4π x 10⁻⁷) x 2000 x 8.0/2π x 40.9 = 0.064 T

Question:

A magnetic field set up using Helmholtz coils (described below) is uniform in a small region and has a magnitude of 0.75T. 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 15kV 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^−1, make a simple guess as to what the beam contains. Why is the answer not unique? Consider two parallel co-axial circular coils of equal radius R, and 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, B=0.72Rμ0​NI​. [Such an arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils].

Answer:

Answer: The beam contains charged particles with a charge-to-mass ratio that is equal to the ratio of the electrostatic field to the magnetic field, i.e. 9.0×10^−5Vm^−1/0.75T = 12.0×10^-6Ckg^-1.

The answer is not unique because different particles can have the same charge-to-mass ratio. For example, a beam of protons and a beam of electrons would both have the same charge-to-mass ratio.

Consider two parallel co-axial circular coils of equal radius R, and 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, B=0.72Rμ0​NI​.

The magnetic field on the axis around the mid-point between the coils can be calculated using Ampere’s law:

B = μ0NI/2πR

Where μ0 is the permeability of free space and I is the current in the coils.

Substituting this into the given equation, we get:

B = 0.72Rμ0NI/2πR

Simplifying this equation, we get:

B = 0.72μ0NI

Which is the given equation.

Question:

A uniform magnetic field of 1.5 T exists in a cylindrical region of radius 10.0 cm, its direction parallel to the axis along east to west. A wire carrying current of 7.0 A 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,(b) the wire is turned from N-S to northeast-northwest direction,(c) the wire in the N-S direction is lowered from the axis by a distance of 6.0 cm?

Answer:

(a) The magnitude of the force on the wire when it intersects the axis is equal to the product of the current and the magnetic field, which is 10.5 N, and the direction of the force is perpendicular to the current and the magnetic field, which is east to west.

(b) The magnitude of the force on the wire when it is turned from N-S to northeast-northwest direction is still 10.5 N, but the direction of the force is now at an angle of 45 degrees from the east-west direction.

(c) The magnitude of the force on the wire when it is lowered from the axis by a distance of 6.0 cm is still 10.5 N, but the direction of the force is now perpendicular to the wire and is pointing downwards.

Question:

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(a) total torque on the coil,(b) the total force on the coil,(c) average force on each electron in the coil due to the magnetic field? (The coil is made of copper wire of cross-sectional area 10^−5m^2, and the free electron density in copper is given to be about 1029 m^−3.)

Answer:

a) Total torque on the coil = (20 x 10 x 0.10 x 5.0) Nm

b) Total force on the coil = (20 x 10 x 0.10 x 5.0) N

c) Average force on each electron in the coil due to the magnetic field = (20 x 10 x 0.10 x 5.0) / (10^−5 x 1029) N

Question:

A solenoid 60 cm long and of radius 4 cm has 3 layers of windings of 300 turns each. A 2.0 cm long wire of mass 2.5 gm 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 6.0 A 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^−2)

Answer:

1. Calculate the magnetic field due to the current in the wire: B = μ0I/2πr = (4π x 10^-7)(6.0 A)/(2π x 0.02 m) = 12.57 x 10^-4 T

2. Calculate the force on the wire due to the magnetic field: F = BIL = (12.57 x 10^-4 T)(6.0 A)(2.0 cm) = 0.152 N

3. Calculate the current in the solenoid windings required to support the weight of the wire: F = mg = (2.5 g)(9.8 ms^-2) = 0.245 N I = F/(BL) = (0.245 N)/[(12.57 x 10^-4 T)(2π x 0.06 m)] = 0.1 A