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A long solenoid of radius R carries a time (t) - dependent current I(t) = I0t(1 - t). A ring of radius 2R is placed co-axially near its middle. During the time interval 0 ≤ t ≤ 1, the induced current (IR) and the induced EMF(VR) in the ring change as
- a)direction of IR remains unchanged and VR is zero at t = 0.5
- b)direction of IR remains unchanged and V, is maximum at t = 0.5
- c)at t = 0.25, direction of IR reverses and VR is maximum
- d)at t = 0.5, direction of IR reverses and VR is zero
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
A long solenoid of radius R carries a time (t) - dependent current I(t...
B = μ0 nl0(t)(1 - t)
Using B0 = μ0 nl0
So, B = B0t(1 - t)
Considering solenoid as an ideal solenoid, extended up to infinity and ring as its centre.
ϕ = BAS = B0t(1 - t)AS
Since induced emf is changing, so current will also be changing because i = e/R
Since direction of emf is changing, so direction of current is also changing.
It will be zero at t = 0.5 sec
Using B0 = μ0 nl0
So, B = B0t(1 - t)
Considering solenoid as an ideal solenoid, extended up to infinity and ring as its centre.
ϕ = BAS = B0t(1 - t)AS
Since induced emf is changing, so current will also be changing because i = e/R
Since direction of emf is changing, so direction of current is also changing.
It will be zero at t = 0.5 sec
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Community Answer
A long solenoid of radius R carries a time (t) - dependent current I(t...
To find the mutual inductance between the solenoid and the ring, we can use the formula for mutual inductance:
M = μ0N1N2A / l
Where:
- M is the mutual inductance
- μ0 is the permeability of free space (4π × 10^−7 T m/A)
- N1 and N2 are the number of turns in the solenoid and the ring, respectively
- A is the cross-sectional area of the solenoid
- l is the length of the solenoid
First, let's find the number of turns in the solenoid. The solenoid is long, so we can consider it as an ideal solenoid with uniform current density. The total number of turns in the solenoid can be calculated using the formula:
N1 = n1 * l1
Where:
- n1 is the number of turns per unit length of the solenoid
- l1 is the length of the solenoid
The current density in the solenoid is given by:
J(t) = I(t) / A
Where:
- I(t) is the time-dependent current in the solenoid
- A is the cross-sectional area of the solenoid
To find n1, we divide J(t) by the current per unit length:
n1 = J(t) / J0
Where:
- J0 is the current per unit length
By substituting I(t) = I0t(1 - t) into J(t), we get:
J(t) = I0t(1 - t) / A
Next, we need to find the current per unit length J0. The total current I flowing through the solenoid is given by:
I = ∫ J(t) dA
Where:
- dA is the differential area element of the cross-section of the solenoid
By substituting J(t) = I0t(1 - t) / A, we can solve for I0:
I = ∫ I0t(1 - t) / A dA
Since the solenoid is long, the integral can be approximated as:
I = I0 ∫ t(1 - t) dA / A
Since the cross-sectional area of the solenoid is constant, we can take it out of the integral:
I = I0 A ∫ t(1 - t) dt
I = I0 A [t^2/2 - t^3/3]
To find the current per unit length J0, we divide I by the length of the solenoid:
J0 = I / l
J0 = I0 A [t^2/2 - t^3/3] / l
Now that we have J0, we can substitute it back into the equation for n1:
n1 = J(t) / J0
n1 = (I0t(1 - t) / A) / [I0 A (t^2/2 - t^3/3) / l]
Simplifying the expression:
n1 = l / A (1 - t) / (t^2/2 - t^3/3)
Next, we can calculate the number of turns in the solenoid N1 by substituting n1 and l
M = μ0N1N2A / l
Where:
- M is the mutual inductance
- μ0 is the permeability of free space (4π × 10^−7 T m/A)
- N1 and N2 are the number of turns in the solenoid and the ring, respectively
- A is the cross-sectional area of the solenoid
- l is the length of the solenoid
First, let's find the number of turns in the solenoid. The solenoid is long, so we can consider it as an ideal solenoid with uniform current density. The total number of turns in the solenoid can be calculated using the formula:
N1 = n1 * l1
Where:
- n1 is the number of turns per unit length of the solenoid
- l1 is the length of the solenoid
The current density in the solenoid is given by:
J(t) = I(t) / A
Where:
- I(t) is the time-dependent current in the solenoid
- A is the cross-sectional area of the solenoid
To find n1, we divide J(t) by the current per unit length:
n1 = J(t) / J0
Where:
- J0 is the current per unit length
By substituting I(t) = I0t(1 - t) into J(t), we get:
J(t) = I0t(1 - t) / A
Next, we need to find the current per unit length J0. The total current I flowing through the solenoid is given by:
I = ∫ J(t) dA
Where:
- dA is the differential area element of the cross-section of the solenoid
By substituting J(t) = I0t(1 - t) / A, we can solve for I0:
I = ∫ I0t(1 - t) / A dA
Since the solenoid is long, the integral can be approximated as:
I = I0 ∫ t(1 - t) dA / A
Since the cross-sectional area of the solenoid is constant, we can take it out of the integral:
I = I0 A ∫ t(1 - t) dt
I = I0 A [t^2/2 - t^3/3]
To find the current per unit length J0, we divide I by the length of the solenoid:
J0 = I / l
J0 = I0 A [t^2/2 - t^3/3] / l
Now that we have J0, we can substitute it back into the equation for n1:
n1 = J(t) / J0
n1 = (I0t(1 - t) / A) / [I0 A (t^2/2 - t^3/3) / l]
Simplifying the expression:
n1 = l / A (1 - t) / (t^2/2 - t^3/3)
Next, we can calculate the number of turns in the solenoid N1 by substituting n1 and l
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A long solenoid of radius R carries a time (t) - dependent current I(t) = I0t(1 - t). A ring of radius 2R is placed co-axially near its middle. During the time interval 0 ≤t ≤1, the induced current (IR) and the induced EMF(VR) in the ring change asa)direction of IR remains unchanged and VR is zero at t = 0.5b)direction of IR remains unchanged and V, is maximum at t = 0.5c)at t = 0.25, direction of IR reverses and VR is maximumd)at t = 0.5, direction of IR reverses and VR is zeroCorrect answer is option 'D'. Can you explain this answer?
Question Description
A long solenoid of radius R carries a time (t) - dependent current I(t) = I0t(1 - t). A ring of radius 2R is placed co-axially near its middle. During the time interval 0 ≤t ≤1, the induced current (IR) and the induced EMF(VR) in the ring change asa)direction of IR remains unchanged and VR is zero at t = 0.5b)direction of IR remains unchanged and V, is maximum at t = 0.5c)at t = 0.25, direction of IR reverses and VR is maximumd)at t = 0.5, direction of IR reverses and VR is zeroCorrect answer is option 'D'. 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 long solenoid of radius R carries a time (t) - dependent current I(t) = I0t(1 - t). A ring of radius 2R is placed co-axially near its middle. During the time interval 0 ≤t ≤1, the induced current (IR) and the induced EMF(VR) in the ring change asa)direction of IR remains unchanged and VR is zero at t = 0.5b)direction of IR remains unchanged and V, is maximum at t = 0.5c)at t = 0.25, direction of IR reverses and VR is maximumd)at t = 0.5, direction of IR reverses and VR is zeroCorrect answer is option 'D'. 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 long solenoid of radius R carries a time (t) - dependent current I(t) = I0t(1 - t). A ring of radius 2R is placed co-axially near its middle. During the time interval 0 ≤t ≤1, the induced current (IR) and the induced EMF(VR) in the ring change asa)direction of IR remains unchanged and VR is zero at t = 0.5b)direction of IR remains unchanged and V, is maximum at t = 0.5c)at t = 0.25, direction of IR reverses and VR is maximumd)at t = 0.5, direction of IR reverses and VR is zeroCorrect answer is option 'D'. Can you explain this answer?.
A long solenoid of radius R carries a time (t) - dependent current I(t) = I0t(1 - t). A ring of radius 2R is placed co-axially near its middle. During the time interval 0 ≤t ≤1, the induced current (IR) and the induced EMF(VR) in the ring change asa)direction of IR remains unchanged and VR is zero at t = 0.5b)direction of IR remains unchanged and V, is maximum at t = 0.5c)at t = 0.25, direction of IR reverses and VR is maximumd)at t = 0.5, direction of IR reverses and VR is zeroCorrect answer is option 'D'. 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 long solenoid of radius R carries a time (t) - dependent current I(t) = I0t(1 - t). A ring of radius 2R is placed co-axially near its middle. During the time interval 0 ≤t ≤1, the induced current (IR) and the induced EMF(VR) in the ring change asa)direction of IR remains unchanged and VR is zero at t = 0.5b)direction of IR remains unchanged and V, is maximum at t = 0.5c)at t = 0.25, direction of IR reverses and VR is maximumd)at t = 0.5, direction of IR reverses and VR is zeroCorrect answer is option 'D'. 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 long solenoid of radius R carries a time (t) - dependent current I(t) = I0t(1 - t). A ring of radius 2R is placed co-axially near its middle. During the time interval 0 ≤t ≤1, the induced current (IR) and the induced EMF(VR) in the ring change asa)direction of IR remains unchanged and VR is zero at t = 0.5b)direction of IR remains unchanged and V, is maximum at t = 0.5c)at t = 0.25, direction of IR reverses and VR is maximumd)at t = 0.5, direction of IR reverses and VR is zeroCorrect answer is option 'D'. Can you explain this answer?.
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A long solenoid of radius R carries a time (t) - dependent current I(t) = I0t(1 - t). A ring of radius 2R is placed co-axially near its middle. During the time interval 0 ≤t ≤1, the induced current (IR) and the induced EMF(VR) in the ring change asa)direction of IR remains unchanged and VR is zero at t = 0.5b)direction of IR remains unchanged and V, is maximum at t = 0.5c)at t = 0.25, direction of IR reverses and VR is maximumd)at t = 0.5, direction of IR reverses and VR is zeroCorrect answer is option 'D'. Can you explain this answer?, a detailed solution for A long solenoid of radius R carries a time (t) - dependent current I(t) = I0t(1 - t). A ring of radius 2R is placed co-axially near its middle. During the time interval 0 ≤t ≤1, the induced current (IR) and the induced EMF(VR) in the ring change asa)direction of IR remains unchanged and VR is zero at t = 0.5b)direction of IR remains unchanged and V, is maximum at t = 0.5c)at t = 0.25, direction of IR reverses and VR is maximumd)at t = 0.5, direction of IR reverses and VR is zeroCorrect answer is option 'D'. Can you explain this answer? has been provided alongside types of A long solenoid of radius R carries a time (t) - dependent current I(t) = I0t(1 - t). A ring of radius 2R is placed co-axially near its middle. During the time interval 0 ≤t ≤1, the induced current (IR) and the induced EMF(VR) in the ring change asa)direction of IR remains unchanged and VR is zero at t = 0.5b)direction of IR remains unchanged and V, is maximum at t = 0.5c)at t = 0.25, direction of IR reverses and VR is maximumd)at t = 0.5, direction of IR reverses and VR is zeroCorrect answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A long solenoid of radius R carries a time (t) - dependent current I(t) = I0t(1 - t). A ring of radius 2R is placed co-axially near its middle. During the time interval 0 ≤t ≤1, the induced current (IR) and the induced EMF(VR) in the ring change asa)direction of IR remains unchanged and VR is zero at t = 0.5b)direction of IR remains unchanged and V, is maximum at t = 0.5c)at t = 0.25, direction of IR reverses and VR is maximumd)at t = 0.5, direction of IR reverses and VR is zeroCorrect answer is option 'D'. Can you explain this answer? tests, examples and also practice JEE tests.
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