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At some location on earth the horizontal component of earth's magnetic field is 18 x 10-6 T. At this location, magnetic neeedle of length 0.12 m and pole strength 1.8 Am is suspended from its mid-point using a thread, it makes 45° angle with horizontal in equilibrium. To keep this needle horizontal, the vertical force that should be applied at one of its ends is :
- a)3.6 × 10–5 N
- b)6.5 × 10–5 N
- c)1.3 × 10–5 N
- d)1.8 × 10–5 N
Correct answer is option 'B'. Can you explain this answer?
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Degrees with the horizontal. What is the magnitude and direction of the torque experienced by the magnetic needle?
To calculate the torque experienced by the magnetic needle, we can use the formula:
τ = mBsinθ
Where:
τ = torque
m = magnetic moment (product of pole strength and length of the needle)
B = magnetic field strength
θ = angle between the magnetic field and the needle
Given:
B = 18 x 10^(-6) T
m = 1.8 Am * 0.12 m = 0.216 Am^2
θ = 45 degrees
Using the formula, we can calculate the torque:
τ = (0.216 Am^2) * (18 x 10^(-6) T) * sin(45 degrees)
= 0.216 * 18 * 10^(-6) * sin(45 degrees)
= 0.216 * 18 * 10^(-6) * (√2 / 2)
= 0.216 * 9 * 10^(-6) * (√2)
= 1.944 x 10^(-6) (√2) Nm
Therefore, the magnitude of the torque experienced by the magnetic needle is approximately 1.944 x 10^(-6) (√2) Nm.
To determine the direction of the torque, we can use the right-hand rule. If we point our right thumb in the direction of the magnetic field, and our fingers in the direction of the magnetic moment, then the direction of the torque will be perpendicular to both. In this case, the torque will be into the page.
So, the direction of the torque experienced by the magnetic needle is into the page.
To calculate the torque experienced by the magnetic needle, we can use the formula:
τ = mBsinθ
Where:
τ = torque
m = magnetic moment (product of pole strength and length of the needle)
B = magnetic field strength
θ = angle between the magnetic field and the needle
Given:
B = 18 x 10^(-6) T
m = 1.8 Am * 0.12 m = 0.216 Am^2
θ = 45 degrees
Using the formula, we can calculate the torque:
τ = (0.216 Am^2) * (18 x 10^(-6) T) * sin(45 degrees)
= 0.216 * 18 * 10^(-6) * sin(45 degrees)
= 0.216 * 18 * 10^(-6) * (√2 / 2)
= 0.216 * 9 * 10^(-6) * (√2)
= 1.944 x 10^(-6) (√2) Nm
Therefore, the magnitude of the torque experienced by the magnetic needle is approximately 1.944 x 10^(-6) (√2) Nm.
To determine the direction of the torque, we can use the right-hand rule. If we point our right thumb in the direction of the magnetic field, and our fingers in the direction of the magnetic moment, then the direction of the torque will be perpendicular to both. In this case, the torque will be into the page.
So, the direction of the torque experienced by the magnetic needle is into the page.
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At some location on earth the horizontal component of earth's magnetic field is 18 x 10-6 T. At this location, magnetic neeedle of length 0.12 m and pole strength 1.8 Am is suspended from its mid-point using a thread, it makes 45° angle with horizontal in equilibrium. To keep this needle horizontal, the vertical force that should be applied at one of its ends is :a)3.6 × 10–5 Nb)6.5 × 10–5 Nc)1.3 × 10–5 Nd)1.8 × 10–5 NCorrect answer is option 'B'. Can you explain this answer?
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At some location on earth the horizontal component of earth's magnetic field is 18 x 10-6 T. At this location, magnetic neeedle of length 0.12 m and pole strength 1.8 Am is suspended from its mid-point using a thread, it makes 45° angle with horizontal in equilibrium. To keep this needle horizontal, the vertical force that should be applied at one of its ends is :a)3.6 × 10–5 Nb)6.5 × 10–5 Nc)1.3 × 10–5 Nd)1.8 × 10–5 NCorrect 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 At some location on earth the horizontal component of earth's magnetic field is 18 x 10-6 T. At this location, magnetic neeedle of length 0.12 m and pole strength 1.8 Am is suspended from its mid-point using a thread, it makes 45° angle with horizontal in equilibrium. To keep this needle horizontal, the vertical force that should be applied at one of its ends is :a)3.6 × 10–5 Nb)6.5 × 10–5 Nc)1.3 × 10–5 Nd)1.8 × 10–5 NCorrect 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 At some location on earth the horizontal component of earth's magnetic field is 18 x 10-6 T. At this location, magnetic neeedle of length 0.12 m and pole strength 1.8 Am is suspended from its mid-point using a thread, it makes 45° angle with horizontal in equilibrium. To keep this needle horizontal, the vertical force that should be applied at one of its ends is :a)3.6 × 10–5 Nb)6.5 × 10–5 Nc)1.3 × 10–5 Nd)1.8 × 10–5 NCorrect answer is option 'B'. Can you explain this answer?.
At some location on earth the horizontal component of earth's magnetic field is 18 x 10-6 T. At this location, magnetic neeedle of length 0.12 m and pole strength 1.8 Am is suspended from its mid-point using a thread, it makes 45° angle with horizontal in equilibrium. To keep this needle horizontal, the vertical force that should be applied at one of its ends is :a)3.6 × 10–5 Nb)6.5 × 10–5 Nc)1.3 × 10–5 Nd)1.8 × 10–5 NCorrect 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 At some location on earth the horizontal component of earth's magnetic field is 18 x 10-6 T. At this location, magnetic neeedle of length 0.12 m and pole strength 1.8 Am is suspended from its mid-point using a thread, it makes 45° angle with horizontal in equilibrium. To keep this needle horizontal, the vertical force that should be applied at one of its ends is :a)3.6 × 10–5 Nb)6.5 × 10–5 Nc)1.3 × 10–5 Nd)1.8 × 10–5 NCorrect 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 At some location on earth the horizontal component of earth's magnetic field is 18 x 10-6 T. At this location, magnetic neeedle of length 0.12 m and pole strength 1.8 Am is suspended from its mid-point using a thread, it makes 45° angle with horizontal in equilibrium. To keep this needle horizontal, the vertical force that should be applied at one of its ends is :a)3.6 × 10–5 Nb)6.5 × 10–5 Nc)1.3 × 10–5 Nd)1.8 × 10–5 NCorrect answer is option 'B'. Can you explain this answer?.
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