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`In gravity tree space, a bead of charge 1 u C andmass 3 mg is threaded on a rough rod of frictioncoefficient u =0.3 A magnetic field of magnitude0.2 T exists perpendicular to the rod. The bead isprojected along the rod with a speed of 4m/ s. How much distance (in m) will the bead coverbefore coming to rest ?`?
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`In gravity tree space, a bead of charge 1 u C andmass 3 mg is threade...
Given data:
Steps to solve the problem:
Calculations:
Step 1: Calculate the gravitational force (Fg) acting on the bead:
The gravitational force can be calculated using the formula:
Fg = m × g
where m is the mass of the bead and g is the acceleration due to gravity.
Given: m = 3 × 10-6 kg
g = 9.8 m/s2 (approximate value)
Fg = (3 × 10-6) × 9.8
Fg ≈ 2.94 × 10-5 N
Step 2: Calculate the magnetic force (FB) acting on the bead:
The magnetic force can be calculated using the formula:
FB = q × v × B
where q is the charge of the bead, v is the velocity of the bead, and B is the magnetic field.
Given: q = 1 μC = 1 × 10-6 C
v = 4 m/s
B = 0.2 T
FB = (1 × 10-6) × 4 × 0.2
FB = 8 × 10-7 N
Step 3: Determine the net force (Fnet) acting on the bead:
The net force can be calculated by subtracting the frictional force (Ff) from the sum of the gravitational force and the magnetic force.
The frictional force can be calculated using the formula:
Ff = μ × FN
where μ is the friction coefficient and FN is the normal force.
Since the bead is projected along the rod, the normal force is equal to the gravitational force (FN = Fg).
Ff = μ × FN
F
- Charge of the bead (q): 1 μC
- Mass of the bead (m): 3 mg = 3 × 10-6 kg
- Friction coefficient (μ): 0.3
- Magnetic field (B): 0.2 T
- Initial speed of the bead (v): 4 m/s
Steps to solve the problem:
- Calculate the gravitational force acting on the bead
- Calculate the magnetic force acting on the bead
- Determine the net force acting on the bead
- Use the net force to calculate the acceleration of the bead
- Use the acceleration to calculate the time taken for the bead to come to rest
- Use the time taken to calculate the distance covered by the bead
Calculations:
Step 1: Calculate the gravitational force (Fg) acting on the bead:
The gravitational force can be calculated using the formula:
Fg = m × g
where m is the mass of the bead and g is the acceleration due to gravity.
Given: m = 3 × 10-6 kg
g = 9.8 m/s2 (approximate value)
Fg = (3 × 10-6) × 9.8
Fg ≈ 2.94 × 10-5 N
Step 2: Calculate the magnetic force (FB) acting on the bead:
The magnetic force can be calculated using the formula:
FB = q × v × B
where q is the charge of the bead, v is the velocity of the bead, and B is the magnetic field.
Given: q = 1 μC = 1 × 10-6 C
v = 4 m/s
B = 0.2 T
FB = (1 × 10-6) × 4 × 0.2
FB = 8 × 10-7 N
Step 3: Determine the net force (Fnet) acting on the bead:
The net force can be calculated by subtracting the frictional force (Ff) from the sum of the gravitational force and the magnetic force.
The frictional force can be calculated using the formula:
Ff = μ × FN
where μ is the friction coefficient and FN is the normal force.
Since the bead is projected along the rod, the normal force is equal to the gravitational force (FN = Fg).
Ff = μ × FN
F
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`In gravity tree space, a bead of charge 1 u C andmass 3 mg is threaded on a rough rod of frictioncoefficient u =0.3 A magnetic field of magnitude0.2 T exists perpendicular to the rod. The bead isprojected along the rod with a speed of 4m/ s. How much distance (in m) will the bead coverbefore coming to rest ?`?
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`In gravity tree space, a bead of charge 1 u C andmass 3 mg is threaded on a rough rod of frictioncoefficient u =0.3 A magnetic field of magnitude0.2 T exists perpendicular to the rod. The bead isprojected along the rod with a speed of 4m/ s. How much distance (in m) will the bead coverbefore coming to rest ?`? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about `In gravity tree space, a bead of charge 1 u C andmass 3 mg is threaded on a rough rod of frictioncoefficient u =0.3 A magnetic field of magnitude0.2 T exists perpendicular to the rod. The bead isprojected along the rod with a speed of 4m/ s. How much distance (in m) will the bead coverbefore coming to rest ?`? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for `In gravity tree space, a bead of charge 1 u C andmass 3 mg is threaded on a rough rod of frictioncoefficient u =0.3 A magnetic field of magnitude0.2 T exists perpendicular to the rod. The bead isprojected along the rod with a speed of 4m/ s. How much distance (in m) will the bead coverbefore coming to rest ?`?.
`In gravity tree space, a bead of charge 1 u C andmass 3 mg is threaded on a rough rod of frictioncoefficient u =0.3 A magnetic field of magnitude0.2 T exists perpendicular to the rod. The bead isprojected along the rod with a speed of 4m/ s. How much distance (in m) will the bead coverbefore coming to rest ?`? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about `In gravity tree space, a bead of charge 1 u C andmass 3 mg is threaded on a rough rod of frictioncoefficient u =0.3 A magnetic field of magnitude0.2 T exists perpendicular to the rod. The bead isprojected along the rod with a speed of 4m/ s. How much distance (in m) will the bead coverbefore coming to rest ?`? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for `In gravity tree space, a bead of charge 1 u C andmass 3 mg is threaded on a rough rod of frictioncoefficient u =0.3 A magnetic field of magnitude0.2 T exists perpendicular to the rod. The bead isprojected along the rod with a speed of 4m/ s. How much distance (in m) will the bead coverbefore coming to rest ?`?.
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