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A thin ring of 10 cm radius carries a uniformly distributed charge. The ring rotates at angular speed of 40 pie rad s^-1 about its axis , perpendicular to its plane.If the magnetic at its centre is 3.8x 10^-9,Then the charge carried by the ring is close to?
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A thin ring of 10 cm radius carries a uniformly distributed charge. Th...
Given Data:
- Radius of the ring (r) = 10 cm = 0.1 m
- Angular speed (ω) = 40π rad/s
- Magnetic field at the center (B) = 3.8 x 10^-9 T
Finding the Charge Carried by the Ring:
Step 1: Finding the Current:
- The current (I) in the ring is given by I = ω*q, where q is the charge.
- ω = 40π rad/s
- The circumference of the ring = 2πr
- The time period for one complete rotation (T) = 1/ω
- The charge passing through any point on the ring in one complete rotation = q
- The current in the ring = q/T = q*ω
- I = q*40π
Step 2: Finding the Force Experienced by the Ring:
- The force experienced by the ring is given by F = B*I*2πr, where B is the magnetic field.
- F = 3.8 x 10^-9 * q * 40π * 2π*0.1
Step 3: Equating Magnetic and Centripetal Forces:
- The magnetic force is equal to the centripetal force acting on the ring.
- F = mv^2/r, where v is the linear velocity of the ring.
- v = r*ω
- F = m*r^2*ω^2/r
- Equating the two forces, we get:
3.8 x 10^-9 * q * 40π * 2π*0.1 = m*0.1^2*(40π)^2/0.1
Step 4: Finding the Charge Carried by the Ring:
- Solving the above equation for q, we get:
q = (m*0.1^2*(40π)^2/0.1) / (3.8 x 10^-9 * 40π * 2π*0.1)
Calculating the Charge:
- By substituting the values of m, q, and solving the equation, we can find the charge carried by the ring.
Therefore, the charge carried by the ring is close to the calculated value.
- Radius of the ring (r) = 10 cm = 0.1 m
- Angular speed (ω) = 40π rad/s
- Magnetic field at the center (B) = 3.8 x 10^-9 T
Finding the Charge Carried by the Ring:
Step 1: Finding the Current:
- The current (I) in the ring is given by I = ω*q, where q is the charge.
- ω = 40π rad/s
- The circumference of the ring = 2πr
- The time period for one complete rotation (T) = 1/ω
- The charge passing through any point on the ring in one complete rotation = q
- The current in the ring = q/T = q*ω
- I = q*40π
Step 2: Finding the Force Experienced by the Ring:
- The force experienced by the ring is given by F = B*I*2πr, where B is the magnetic field.
- F = 3.8 x 10^-9 * q * 40π * 2π*0.1
Step 3: Equating Magnetic and Centripetal Forces:
- The magnetic force is equal to the centripetal force acting on the ring.
- F = mv^2/r, where v is the linear velocity of the ring.
- v = r*ω
- F = m*r^2*ω^2/r
- Equating the two forces, we get:
3.8 x 10^-9 * q * 40π * 2π*0.1 = m*0.1^2*(40π)^2/0.1
Step 4: Finding the Charge Carried by the Ring:
- Solving the above equation for q, we get:
q = (m*0.1^2*(40π)^2/0.1) / (3.8 x 10^-9 * 40π * 2π*0.1)
Calculating the Charge:
- By substituting the values of m, q, and solving the equation, we can find the charge carried by the ring.
Therefore, the charge carried by the ring is close to the calculated value.
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A thin ring of 10 cm radius carries a uniformly distributed charge. The ring rotates at angular speed of 40 pie rad s^-1 about its axis , perpendicular to its plane.If the magnetic at its centre is 3.8x 10^-9,Then the charge carried by the ring is close to?
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A thin ring of 10 cm radius carries a uniformly distributed charge. The ring rotates at angular speed of 40 pie rad s^-1 about its axis , perpendicular to its plane.If the magnetic at its centre is 3.8x 10^-9,Then the charge carried by the ring is close to? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A thin ring of 10 cm radius carries a uniformly distributed charge. The ring rotates at angular speed of 40 pie rad s^-1 about its axis , perpendicular to its plane.If the magnetic at its centre is 3.8x 10^-9,Then the charge carried by the ring is close to? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A thin ring of 10 cm radius carries a uniformly distributed charge. The ring rotates at angular speed of 40 pie rad s^-1 about its axis , perpendicular to its plane.If the magnetic at its centre is 3.8x 10^-9,Then the charge carried by the ring is close to?.
A thin ring of 10 cm radius carries a uniformly distributed charge. The ring rotates at angular speed of 40 pie rad s^-1 about its axis , perpendicular to its plane.If the magnetic at its centre is 3.8x 10^-9,Then the charge carried by the ring is close to? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A thin ring of 10 cm radius carries a uniformly distributed charge. The ring rotates at angular speed of 40 pie rad s^-1 about its axis , perpendicular to its plane.If the magnetic at its centre is 3.8x 10^-9,Then the charge carried by the ring is close to? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A thin ring of 10 cm radius carries a uniformly distributed charge. The ring rotates at angular speed of 40 pie rad s^-1 about its axis , perpendicular to its plane.If the magnetic at its centre is 3.8x 10^-9,Then the charge carried by the ring is close to?.
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