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A uniform disc of radius R having charge Q distributed uniformly all over its surface is placed on a smooth horizontal surface. A magnetic field, B = kxt2, where k is a constant, x is the distance (in metre) from the centre of the disc and t is the time (in second), is switched on perpendicular to the plane of the disc. Find the torque (in N-m) acting on the disc after 15 sec. (Take 4kQ = 1 S.I.unit and R = 1m) is
Correct answer is '1'. Can you explain this answer?
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A uniform disc of radius R having charge Q distributed uniformly all o...
Solution:
Given data:
- The radius of the disc, R = 1m
- The charge on the disc, Q = (1/4k) C
- The magnetic field, B = kxt^2 T
- The time for which the magnetic field is switched on, t = 15s
Let's calculate the magnetic moment of the disc first.
Magnetic Moment of the Disc:
- The magnetic moment of the disc is given by the product of the current flowing through it and the area of the disc.
- The current flowing through the disc is due to the induced electric field in it, which is given by E = -dΦ/dt, where Φ is the magnetic flux through the disc.
- The magnetic flux through the disc is given by Φ = BA, where A is the area of the disc and B is the magnetic field.
- So, the induced electric field in the disc is given by E = -d(BA)/dt = -AB(dB/dt) = -Akx^2.
- The current flowing through the disc is given by I = σE, where σ is the surface charge density of the disc.
- The surface charge density of the disc is given by σ = Q/4πR^2.
- So, the current flowing through the disc is given by I = (Q/4πR^2)(-Akx^2).
- The magnetic moment of the disc is given by M = IA = (Q/4πR^2)(-Akx^2)(πR^2) = -AkQx^2.
Let's calculate the torque acting on the disc now.
Torque on the Disc:
- The torque acting on the disc is given by the cross product of the magnetic moment of the disc and the magnetic field.
- So, the torque acting on the disc is given by τ = M × B = (-AkQx^2) × kxt^2.
- The direction of the torque is perpendicular to both the magnetic moment and the magnetic field, which is in the z-direction.
- So, the magnitude of the torque acting on the disc is given by |τ| = Ak^2Qx^4t^2.
- Substituting the given values, we get |τ| = (1/4)(1/16)(1^4)(15^2) = 1 N-m.
Therefore, the torque acting on the disc after 15 sec is 1 N-m.
Given data:
- The radius of the disc, R = 1m
- The charge on the disc, Q = (1/4k) C
- The magnetic field, B = kxt^2 T
- The time for which the magnetic field is switched on, t = 15s
Let's calculate the magnetic moment of the disc first.
Magnetic Moment of the Disc:
- The magnetic moment of the disc is given by the product of the current flowing through it and the area of the disc.
- The current flowing through the disc is due to the induced electric field in it, which is given by E = -dΦ/dt, where Φ is the magnetic flux through the disc.
- The magnetic flux through the disc is given by Φ = BA, where A is the area of the disc and B is the magnetic field.
- So, the induced electric field in the disc is given by E = -d(BA)/dt = -AB(dB/dt) = -Akx^2.
- The current flowing through the disc is given by I = σE, where σ is the surface charge density of the disc.
- The surface charge density of the disc is given by σ = Q/4πR^2.
- So, the current flowing through the disc is given by I = (Q/4πR^2)(-Akx^2).
- The magnetic moment of the disc is given by M = IA = (Q/4πR^2)(-Akx^2)(πR^2) = -AkQx^2.
Let's calculate the torque acting on the disc now.
Torque on the Disc:
- The torque acting on the disc is given by the cross product of the magnetic moment of the disc and the magnetic field.
- So, the torque acting on the disc is given by τ = M × B = (-AkQx^2) × kxt^2.
- The direction of the torque is perpendicular to both the magnetic moment and the magnetic field, which is in the z-direction.
- So, the magnitude of the torque acting on the disc is given by |τ| = Ak^2Qx^4t^2.
- Substituting the given values, we get |τ| = (1/4)(1/16)(1^4)(15^2) = 1 N-m.
Therefore, the torque acting on the disc after 15 sec is 1 N-m.
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A uniform disc of radius R having charge Q distributed uniformly all over its surface is placed on a smooth horizontal surface. A magnetic field, B = kxt2, where k is a constant, x is the distance (in metre) from the centre of the disc and t is the time (in second), is switched on perpendicular to the plane of the disc. Find the torque (in N-m) acting on the disc after 15 sec. (Take 4kQ = 1 S.I.unit and R = 1m) isCorrect answer is '1'. Can you explain this answer?
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A uniform disc of radius R having charge Q distributed uniformly all over its surface is placed on a smooth horizontal surface. A magnetic field, B = kxt2, where k is a constant, x is the distance (in metre) from the centre of the disc and t is the time (in second), is switched on perpendicular to the plane of the disc. Find the torque (in N-m) acting on the disc after 15 sec. (Take 4kQ = 1 S.I.unit and R = 1m) isCorrect answer is '1'. 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 disc of radius R having charge Q distributed uniformly all over its surface is placed on a smooth horizontal surface. A magnetic field, B = kxt2, where k is a constant, x is the distance (in metre) from the centre of the disc and t is the time (in second), is switched on perpendicular to the plane of the disc. Find the torque (in N-m) acting on the disc after 15 sec. (Take 4kQ = 1 S.I.unit and R = 1m) isCorrect answer is '1'. 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 disc of radius R having charge Q distributed uniformly all over its surface is placed on a smooth horizontal surface. A magnetic field, B = kxt2, where k is a constant, x is the distance (in metre) from the centre of the disc and t is the time (in second), is switched on perpendicular to the plane of the disc. Find the torque (in N-m) acting on the disc after 15 sec. (Take 4kQ = 1 S.I.unit and R = 1m) isCorrect answer is '1'. Can you explain this answer?.
A uniform disc of radius R having charge Q distributed uniformly all over its surface is placed on a smooth horizontal surface. A magnetic field, B = kxt2, where k is a constant, x is the distance (in metre) from the centre of the disc and t is the time (in second), is switched on perpendicular to the plane of the disc. Find the torque (in N-m) acting on the disc after 15 sec. (Take 4kQ = 1 S.I.unit and R = 1m) isCorrect answer is '1'. 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 disc of radius R having charge Q distributed uniformly all over its surface is placed on a smooth horizontal surface. A magnetic field, B = kxt2, where k is a constant, x is the distance (in metre) from the centre of the disc and t is the time (in second), is switched on perpendicular to the plane of the disc. Find the torque (in N-m) acting on the disc after 15 sec. (Take 4kQ = 1 S.I.unit and R = 1m) isCorrect answer is '1'. 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 disc of radius R having charge Q distributed uniformly all over its surface is placed on a smooth horizontal surface. A magnetic field, B = kxt2, where k is a constant, x is the distance (in metre) from the centre of the disc and t is the time (in second), is switched on perpendicular to the plane of the disc. Find the torque (in N-m) acting on the disc after 15 sec. (Take 4kQ = 1 S.I.unit and R = 1m) isCorrect answer is '1'. Can you explain this answer?.
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