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A uniform magnetic needle is suspended from its centre by a thread. Its upper end is now loaded with a mass 50 mg , and the needle becomes horizontal. If the strength of each pole is 98.1 A cm and g = 981 cm s−2, then the vertical component of earth's magnetic induction is
- a)0.50 G
- b)0.25 G
- c)0.005 G
- d)0.05 G
Correct answer is option 'B'. Can you explain this answer?
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A uniform magnetic needle is suspended from its centre by a thread. It...
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, find the magnetic moment of the needle.
We can start by using the equation for the torque on a magnetic dipole in a uniform magnetic field:
τ = m × B
where τ is the torque, m is the magnetic moment, and B is the magnetic field. Since the needle is horizontal, we know that the torque must be zero, which means that the magnetic moment and the magnetic field must be perpendicular to each other.
We can also use the equation for the torque due to a weight hanging from the needle:
τ = r × F
where r is the distance from the pivot point to the point where the force is applied (in this case, the centre of mass of the needle), and F is the force (in this case, the weight of the load).
Since the needle is horizontal, we know that the torque due to the weight must be equal and opposite to the torque due to the magnetic field, so we can set the two equations equal to each other:
m × B = r × F
We can rearrange this equation to solve for the magnetic moment:
m = (r × F) / B
To find r, we need to know the length of the needle. Let's assume that the needle is 10 cm long. Then the distance from the pivot point to the centre of mass is 5 cm.
To find the magnetic field, we can use the equation for the magnetic field due to a pole:
B = μ0I / (2πr)
where μ0 is the permeability of free space, I is the current (which is the same for both poles), and r is the distance from the pole.
Since the needle is uniform, we can assume that the magnetic field at the centre of the needle is the same as the magnetic field at the pole. Let's assume that the current is 1 A. Then the magnetic field at the centre of the needle is:
B = μ0I / (2πr) = (4π × 10-7 T m/A) × 1 A / (2π × 0.05 m) = 1.26 × 10-4 T
Now we can plug in all the values and solve for the magnetic moment:
m = (r × F) / B = (0.05 m × 50 × 10-6 kg × 9.81 m/s2) / (1.26 × 10-4 T) = 1.92 × 10-6 J/T
Therefore, the magnetic moment of the needle is 1.92 × 10-6 J/T.
, find the magnetic moment of the needle.
We can start by using the equation for the torque on a magnetic dipole in a uniform magnetic field:
τ = m × B
where τ is the torque, m is the magnetic moment, and B is the magnetic field. Since the needle is horizontal, we know that the torque must be zero, which means that the magnetic moment and the magnetic field must be perpendicular to each other.
We can also use the equation for the torque due to a weight hanging from the needle:
τ = r × F
where r is the distance from the pivot point to the point where the force is applied (in this case, the centre of mass of the needle), and F is the force (in this case, the weight of the load).
Since the needle is horizontal, we know that the torque due to the weight must be equal and opposite to the torque due to the magnetic field, so we can set the two equations equal to each other:
m × B = r × F
We can rearrange this equation to solve for the magnetic moment:
m = (r × F) / B
To find r, we need to know the length of the needle. Let's assume that the needle is 10 cm long. Then the distance from the pivot point to the centre of mass is 5 cm.
To find the magnetic field, we can use the equation for the magnetic field due to a pole:
B = μ0I / (2πr)
where μ0 is the permeability of free space, I is the current (which is the same for both poles), and r is the distance from the pole.
Since the needle is uniform, we can assume that the magnetic field at the centre of the needle is the same as the magnetic field at the pole. Let's assume that the current is 1 A. Then the magnetic field at the centre of the needle is:
B = μ0I / (2πr) = (4π × 10-7 T m/A) × 1 A / (2π × 0.05 m) = 1.26 × 10-4 T
Now we can plug in all the values and solve for the magnetic moment:
m = (r × F) / B = (0.05 m × 50 × 10-6 kg × 9.81 m/s2) / (1.26 × 10-4 T) = 1.92 × 10-6 J/T
Therefore, the magnetic moment of the needle is 1.92 × 10-6 J/T.
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Community Answer
A uniform magnetic needle is suspended from its centre by a thread. It...
As shown in figure,
mass M = 50mg = 50 × 10−3g
Strength of each pole,
m = 98.1 amp cm , g = 981 cm s−2, V = ?
In equilibrium, mV × 2l = Mg × l
mass M = 50mg = 50 × 10−3g
Strength of each pole,
m = 98.1 amp cm , g = 981 cm s−2, V = ?
In equilibrium, mV × 2l = Mg × l
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A uniform magnetic needle is suspended from its centre by a thread. Its upper end is now loaded with a mass 50 mg , and the needle becomes horizontal. If the strength of each pole is 98.1 A cmand g = 981 cm s−2, then the vertical component of earths magnetic induction isa)0.50 Gb)0.25 Gc)0.005 Gd)0.05 GCorrect answer is option 'B'. Can you explain this answer?
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A uniform magnetic needle is suspended from its centre by a thread. Its upper end is now loaded with a mass 50 mg , and the needle becomes horizontal. If the strength of each pole is 98.1 A cmand g = 981 cm s−2, then the vertical component of earths magnetic induction isa)0.50 Gb)0.25 Gc)0.005 Gd)0.05 GCorrect answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A uniform magnetic needle is suspended from its centre by a thread. Its upper end is now loaded with a mass 50 mg , and the needle becomes horizontal. If the strength of each pole is 98.1 A cmand g = 981 cm s−2, then the vertical component of earths magnetic induction isa)0.50 Gb)0.25 Gc)0.005 Gd)0.05 GCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A uniform magnetic needle is suspended from its centre by a thread. Its upper end is now loaded with a mass 50 mg , and the needle becomes horizontal. If the strength of each pole is 98.1 A cmand g = 981 cm s−2, then the vertical component of earths magnetic induction isa)0.50 Gb)0.25 Gc)0.005 Gd)0.05 GCorrect answer is option 'B'. Can you explain this answer?.
A uniform magnetic needle is suspended from its centre by a thread. Its upper end is now loaded with a mass 50 mg , and the needle becomes horizontal. If the strength of each pole is 98.1 A cmand g = 981 cm s−2, then the vertical component of earths magnetic induction isa)0.50 Gb)0.25 Gc)0.005 Gd)0.05 GCorrect answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A uniform magnetic needle is suspended from its centre by a thread. Its upper end is now loaded with a mass 50 mg , and the needle becomes horizontal. If the strength of each pole is 98.1 A cmand g = 981 cm s−2, then the vertical component of earths magnetic induction isa)0.50 Gb)0.25 Gc)0.005 Gd)0.05 GCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A uniform magnetic needle is suspended from its centre by a thread. Its upper end is now loaded with a mass 50 mg , and the needle becomes horizontal. If the strength of each pole is 98.1 A cmand g = 981 cm s−2, then the vertical component of earths magnetic induction isa)0.50 Gb)0.25 Gc)0.005 Gd)0.05 GCorrect answer is option 'B'. Can you explain this answer?.
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