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A rectangular loop of wire in free space joins points A(1, 0, 1) to B(3, 0, 1) to C(3, 0, 4) to D(1, 0, 4) to A. The wire carries a current of 6 mA flowing in the uz direction from B to C. A filamentary current of 15 A flows along the entire z, axis in the uz directions.
Que: The force on side BC is
- a)-18ux nN
- b)18ux nN
- c)3.6 ux nN
- d)-3.6 ux nN
Correct answer is option 'A'. Can you explain this answer?
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A rectangular loop of wire in free space joins points A(1, 0, 1) to B(...
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A rectangular loop of wire in free space joins points A(1, 0, 1) to B(...
Given:
- Rectangular loop of wire joining points A(1,0,1) to B(3,0,1) to C(3,0,4) to D(1,0,4) to A
- Current of 6 mA flowing in the uz direction from B to C
- Filamentary current of 15 A flows along the entire z-axis in the uz direction
To find:
- Force on side BC
Solution:
1. Find the magnetic field due to the filamentary current along the z-axis at point B and C.
- The magnetic field at point P due to current I flowing through a straight wire located a distance r away from P is given by:
B = (μ₀ / 4π) * I / r
- At point B, the distance from the filamentary current is r = 3 m.
- Therefore, the magnetic field at point B is:
B_B = (μ₀ / 4π) * 15 / 3 = 5μ₀ / 4π T
- At point C, the distance from the filamentary current is r = 1 m.
- Therefore, the magnetic field at point C is:
B_C = (μ₀ / 4π) * 15 / 1 = 15μ₀ / 4π T
2. Find the magnetic field due to the current in the rectangular loop at points B and C.
- The magnetic field at point P due to a current element Idl is given by:
d𝐵 = (μ₀ / 4π) * Idl x ẑ / r²
- Consider the current flowing from B to C. The current element dl is in the negative x direction.
- At point B, the current element is perpendicular to the line joining B and C, so the magnetic field due to this element is zero.
- At point C, the current element is in the positive x direction, so the magnetic field due to this element is:
d𝐵_C = (μ₀ / 4π) * 6e-3 * (-dx) * ẑ / r², where dx is the length of the current element
= -(μ₀ / 4π) * 6e-3 * dx * ẑ / 9
- The total magnetic field at point C due to the current in the rectangular loop is obtained by integrating over the loop:
B_C' = ∫d𝐵_C
= -(μ₀ / 4π) * 6e-3 * ∫dx / 9 * ∫dy / (4 - 1) * ∫dz / (3 - 1) ẑ
= -(μ₀ / 4π) * 6e-3 * (2/3) * ẑ
= -4μ₀ / 1000 * ẑ T
3. Find the force on side BC using the Lorentz force law.
- The force on a current element Idl due to a magnetic field B is given by:
dF = Idl x B
- The total force on a current-carrying loop due to a magnetic field B is obtained by integrating over the loop:
F = ∫Idl x B
- Consider the current flowing from B
- Rectangular loop of wire joining points A(1,0,1) to B(3,0,1) to C(3,0,4) to D(1,0,4) to A
- Current of 6 mA flowing in the uz direction from B to C
- Filamentary current of 15 A flows along the entire z-axis in the uz direction
To find:
- Force on side BC
Solution:
1. Find the magnetic field due to the filamentary current along the z-axis at point B and C.
- The magnetic field at point P due to current I flowing through a straight wire located a distance r away from P is given by:
B = (μ₀ / 4π) * I / r
- At point B, the distance from the filamentary current is r = 3 m.
- Therefore, the magnetic field at point B is:
B_B = (μ₀ / 4π) * 15 / 3 = 5μ₀ / 4π T
- At point C, the distance from the filamentary current is r = 1 m.
- Therefore, the magnetic field at point C is:
B_C = (μ₀ / 4π) * 15 / 1 = 15μ₀ / 4π T
2. Find the magnetic field due to the current in the rectangular loop at points B and C.
- The magnetic field at point P due to a current element Idl is given by:
d𝐵 = (μ₀ / 4π) * Idl x ẑ / r²
- Consider the current flowing from B to C. The current element dl is in the negative x direction.
- At point B, the current element is perpendicular to the line joining B and C, so the magnetic field due to this element is zero.
- At point C, the current element is in the positive x direction, so the magnetic field due to this element is:
d𝐵_C = (μ₀ / 4π) * 6e-3 * (-dx) * ẑ / r², where dx is the length of the current element
= -(μ₀ / 4π) * 6e-3 * dx * ẑ / 9
- The total magnetic field at point C due to the current in the rectangular loop is obtained by integrating over the loop:
B_C' = ∫d𝐵_C
= -(μ₀ / 4π) * 6e-3 * ∫dx / 9 * ∫dy / (4 - 1) * ∫dz / (3 - 1) ẑ
= -(μ₀ / 4π) * 6e-3 * (2/3) * ẑ
= -4μ₀ / 1000 * ẑ T
3. Find the force on side BC using the Lorentz force law.
- The force on a current element Idl due to a magnetic field B is given by:
dF = Idl x B
- The total force on a current-carrying loop due to a magnetic field B is obtained by integrating over the loop:
F = ∫Idl x B
- Consider the current flowing from B
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A rectangular loop of wire in free space joins points A(1, 0, 1) to B(3, 0, 1) to C(3, 0, 4) to D(1, 0, 4) to A. The wire carries a current of 6 mA flowing in the uzdirection from B to C. A filamentary current of 15 A flows along the entire z, axis in the uzdirections.Que:The force on side BC isa)-18uxnNb)18uxnNc)3.6uxnNd)-3.6uxnNCorrect answer is option 'A'. Can you explain this answer?
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