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The magnetic field due to a current carrying circular loop of radius 3 cm at a point on the axis at a distance of 4 cm from the centre is 54 μT. What will be its value at the centre of loop?
- a)125 μT
- b)150 μT
- c)250 μT
- d)75 μT
Correct answer is option 'C'. Can you explain this answer?
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The magnetic field due to a current carrying circular loop of radius 3...
The magnetic field at a point on the axis of a circular loop at a distance x from centre is,
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The magnetic field due to a current carrying circular loop of radius 3...
Given Data:
- Radius of the circular loop, r = 3 cm
- Distance from the centre of the loop to the point on the axis, x = 4 cm
- Magnetic field at the point on the axis, B = 54 μT
Formula for Magnetic Field on the Axis of Circular Current Loop:
The magnetic field at a point on the axis of a circular current loop is given by the formula:
\[ B = \frac{\mu_0 I r^2}{2(x^2 + r^2)^{3/2}} \]
Calculation for Magnetic Field at the Centre of the Loop:
- At the centre of the loop, x = 0
- Substituting x = 0 in the formula, we get:
\[ B_{\text{centre}} = \frac{\mu_0 I r^2}{2r^3} = \frac{\mu_0 I}{2r} \]
- Given that B = 54 μT at a point on the axis, x = 4 cm, we can calculate the current I using the formula:
\[ 54 \times 10^{-6} = \frac{4\pi \times 10^{-7} \times I \times (3 \times 10^{-2})^2}{2((4 \times 10^{-2})^2 + (3 \times 10^{-2})^2)^{3/2}} \]
- Solving for I, we get:
\[ I = \frac{54 \times 10^{-6} \times 2 \times ((4 \times 10^{-2})^2 + (3 \times 10^{-2})^2)^{3/2}}{4\pi \times 10^{-7} \times (3 \times 10^{-2})^2} \]
- Once we have the value of I, we can calculate the magnetic field at the centre of the loop using the formula:
\[ B_{\text{centre}} = \frac{\mu_0 I}{2r} \]
Therefore, the magnetic field at the centre of the loop will be 250 μT (option 'C').
- Radius of the circular loop, r = 3 cm
- Distance from the centre of the loop to the point on the axis, x = 4 cm
- Magnetic field at the point on the axis, B = 54 μT
Formula for Magnetic Field on the Axis of Circular Current Loop:
The magnetic field at a point on the axis of a circular current loop is given by the formula:
\[ B = \frac{\mu_0 I r^2}{2(x^2 + r^2)^{3/2}} \]
Calculation for Magnetic Field at the Centre of the Loop:
- At the centre of the loop, x = 0
- Substituting x = 0 in the formula, we get:
\[ B_{\text{centre}} = \frac{\mu_0 I r^2}{2r^3} = \frac{\mu_0 I}{2r} \]
- Given that B = 54 μT at a point on the axis, x = 4 cm, we can calculate the current I using the formula:
\[ 54 \times 10^{-6} = \frac{4\pi \times 10^{-7} \times I \times (3 \times 10^{-2})^2}{2((4 \times 10^{-2})^2 + (3 \times 10^{-2})^2)^{3/2}} \]
- Solving for I, we get:
\[ I = \frac{54 \times 10^{-6} \times 2 \times ((4 \times 10^{-2})^2 + (3 \times 10^{-2})^2)^{3/2}}{4\pi \times 10^{-7} \times (3 \times 10^{-2})^2} \]
- Once we have the value of I, we can calculate the magnetic field at the centre of the loop using the formula:
\[ B_{\text{centre}} = \frac{\mu_0 I}{2r} \]
Therefore, the magnetic field at the centre of the loop will be 250 μT (option 'C').
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The magnetic field due to a current carrying circular loop of radius 3 cm at a point on the axis at a distance of 4 cm from the centre is 54 μT. What will be its value at the centre of loop?a)125 μTb)150 μTc)250 μTd)75 μTCorrect answer is option 'C'. Can you explain this answer?
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The magnetic field due to a current carrying circular loop of radius 3 cm at a point on the axis at a distance of 4 cm from the centre is 54 μT. What will be its value at the centre of loop?a)125 μTb)150 μTc)250 μTd)75 μTCorrect answer is option 'C'. 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 The magnetic field due to a current carrying circular loop of radius 3 cm at a point on the axis at a distance of 4 cm from the centre is 54 μT. What will be its value at the centre of loop?a)125 μTb)150 μTc)250 μTd)75 μTCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The magnetic field due to a current carrying circular loop of radius 3 cm at a point on the axis at a distance of 4 cm from the centre is 54 μT. What will be its value at the centre of loop?a)125 μTb)150 μTc)250 μTd)75 μTCorrect answer is option 'C'. Can you explain this answer?.
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