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A magnet of length 10cm and magnetic moment 1 Am2 is placed along the side AB of an equilateral triangle ABC. If the length of the side AB is 10cm, the magnetic induction at the point 'C' is ( μ0 = 4 π x 10-7 Hm-1 )
- a)10-9T
- b)10-7T
- c)10-5T
- d)10-4T
Correct answer is option 'D'. Can you explain this answer?
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A magnet of length 10cm and magnetic moment 1 Am2 is placed along the ...
Assuming that the equilateral triangle ABC lies in a plane, we can use the formula for the magnetic field at a point on the axis of a magnetic dipole:
B = μ0/(4π) * (2m/r^3) * (cosθ1 - cosθ2)
where μ0 is the permeability of free space, m is the magnetic moment of the magnet, r is the distance from the magnet to the point of interest, θ1 is the angle between the axis of the magnet and the line from the magnet to the point of interest, and θ2 is the angle between the axis of the magnet and the line from the magnet to the point on the opposite side of the magnet.
In this case, we can choose the x-axis to be along the line AB, with the origin at point A. The magnet is then aligned along the x-axis, with its center at x = 5 cm. The point of interest is point C, which lies on the y-axis at a distance of 5√3 cm from the origin. The angle θ1 between the axis of the magnet and the line from the magnet to point C is 60 degrees, and the angle θ2 between the axis of the magnet and the line from the magnet to the point on the opposite side of the magnet (point D) is also 60 degrees. The distance from the magnet to point C is r = 5√7 cm.
Plugging in these values, we get:
B = μ0/(4π) * (2 * 1 Am2 / (5√7 cm)^3) * (cos60° - cos60°) = μ0/(4π) * (2 * 1 Am2 / (5√7 cm)^3) * 0 = 0
Therefore, the magnetic induction at point C is zero. This makes sense intuitively, since the magnetic field lines from the magnet are symmetrically distributed around the x-axis, and point C lies on the y-axis.
B = μ0/(4π) * (2m/r^3) * (cosθ1 - cosθ2)
where μ0 is the permeability of free space, m is the magnetic moment of the magnet, r is the distance from the magnet to the point of interest, θ1 is the angle between the axis of the magnet and the line from the magnet to the point of interest, and θ2 is the angle between the axis of the magnet and the line from the magnet to the point on the opposite side of the magnet.
In this case, we can choose the x-axis to be along the line AB, with the origin at point A. The magnet is then aligned along the x-axis, with its center at x = 5 cm. The point of interest is point C, which lies on the y-axis at a distance of 5√3 cm from the origin. The angle θ1 between the axis of the magnet and the line from the magnet to point C is 60 degrees, and the angle θ2 between the axis of the magnet and the line from the magnet to the point on the opposite side of the magnet (point D) is also 60 degrees. The distance from the magnet to point C is r = 5√7 cm.
Plugging in these values, we get:
B = μ0/(4π) * (2 * 1 Am2 / (5√7 cm)^3) * (cos60° - cos60°) = μ0/(4π) * (2 * 1 Am2 / (5√7 cm)^3) * 0 = 0
Therefore, the magnetic induction at point C is zero. This makes sense intuitively, since the magnetic field lines from the magnet are symmetrically distributed around the x-axis, and point C lies on the y-axis.
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A magnet of length 10cm and magnetic moment 1 Am2 is placed along the side AB of an equilateral triangle ABC. If the length of the side AB is 10cm, the magnetic induction at the point 'C' is ( μ0= 4 π x10-7 Hm-1 )a)10-9Tb)10-7Tc)10-5Td)10-4TCorrect answer is option 'D'. Can you explain this answer?
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A magnet of length 10cm and magnetic moment 1 Am2 is placed along the side AB of an equilateral triangle ABC. If the length of the side AB is 10cm, the magnetic induction at the point 'C' is ( μ0= 4 π x10-7 Hm-1 )a)10-9Tb)10-7Tc)10-5Td)10-4TCorrect answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A magnet of length 10cm and magnetic moment 1 Am2 is placed along the side AB of an equilateral triangle ABC. If the length of the side AB is 10cm, the magnetic induction at the point 'C' is ( μ0= 4 π x10-7 Hm-1 )a)10-9Tb)10-7Tc)10-5Td)10-4TCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A magnet of length 10cm and magnetic moment 1 Am2 is placed along the side AB of an equilateral triangle ABC. If the length of the side AB is 10cm, the magnetic induction at the point 'C' is ( μ0= 4 π x10-7 Hm-1 )a)10-9Tb)10-7Tc)10-5Td)10-4TCorrect answer is option 'D'. Can you explain this answer?.
A magnet of length 10cm and magnetic moment 1 Am2 is placed along the side AB of an equilateral triangle ABC. If the length of the side AB is 10cm, the magnetic induction at the point 'C' is ( μ0= 4 π x10-7 Hm-1 )a)10-9Tb)10-7Tc)10-5Td)10-4TCorrect answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A magnet of length 10cm and magnetic moment 1 Am2 is placed along the side AB of an equilateral triangle ABC. If the length of the side AB is 10cm, the magnetic induction at the point 'C' is ( μ0= 4 π x10-7 Hm-1 )a)10-9Tb)10-7Tc)10-5Td)10-4TCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A magnet of length 10cm and magnetic moment 1 Am2 is placed along the side AB of an equilateral triangle ABC. If the length of the side AB is 10cm, the magnetic induction at the point 'C' is ( μ0= 4 π x10-7 Hm-1 )a)10-9Tb)10-7Tc)10-5Td)10-4TCorrect answer is option 'D'. Can you explain this answer?.
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