Let us consider the following MCQ which appeared in AIEEE 2004 question paper:
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 centro of the loop?
(a) 250 μT (b) 150 μT (c) 125 μT (d) 75 μT
The magnetic flux density(B) at a point on the axis at distance ‘x’ from the centre for a single turn loop is given by
B = μ0 r2I / [2(r²+ x²)3/2] where μ0 is the permeabiloty of free space, ‘r’ is the radius of the loop and ‘I’ is the current.
The magnetic field(B') at the centre of the loop is given by
B' = μ0 I/2r
Dividing, B/B' = r3/(r²+ x²)3/2 from which B' = B[(r²+ x²)3/2] /r3.
You may substitute distances in cm itself since their units will get cancelled in the ratio. On substituting B in μT itself, you will get the answer in μT.
Therefore, B' = 54×[(9+16)3/2]/27 = 54×125/27 = 250 μT.
The following multiple choice question appeared in Kerala Medical Entrance 2003 test paper:
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 centro of the loop?
(a) 250 μT (b) 150 μT (c) 125 μT (d) 75 μT
The magnetic flux density(B) at a point on the axis at distance ‘x’ from the centre for a single turn loop is given by
B = μ0 r2I / [2(r²+ x²)3/2] where μ0 is the permeabiloty of free space, ‘r’ is the radius of the loop and ‘I’ is the current.
The magnetic field(B') at the centre of the loop is given by
B' = μ0 I/2r
Dividing, B/B' = r3/(r²+ x²)3/2 from which B' = B[(r²+ x²)3/2] /r3.
You may substitute distances in cm itself since their units will get cancelled in the ratio. On substituting B in μT itself, you will get the answer in μT.
Therefore, B' = 54×[(9+16)3/2]/27 = 54×125/27 = 250 μT.
The following multiple choice question appeared in Kerala Medical Entrance 2003 test paper:
A wire of certain length carries a steady current. It is first bent to form a circular coil of one turn. The same wire is next bent to form a circular coil of three turns. The ratio of magnetic induction at the centre of the coil in the two cases is
(a) 9:1 (b) 1:9 (c) 1:3 (d) 3:1 (e) 1:1
(a) 9:1 (b) 1:9 (c) 1:3 (d) 3:1 (e) 1:1
Since the magnetic induction (flux density) at the centre of a circular coil is given by
B = μ0 nI/(2r) where μ0 is the permeability of free space, ‘n’ is the number of turns in the coil, I is the current and ‘r’ is the radius of the coil, we have, B α n/r for a given current.
B = μ0 nI/(2r) where μ0 is the permeability of free space, ‘n’ is the number of turns in the coil, I is the current and ‘r’ is the radius of the coil, we have, B α n/r for a given current.
The ratio of magnetic inductions at the centre is therefore B/B' = (n/n') ×(r'/r) = (1/3) ×(1/3) = 1/9, since 2πr = 3×2πr' so that r'/r = 1/3. The correct option therefore is (b).
Here is a simple question (but be careful):
Here is a simple question (but be careful):
An infinitely long aluminium pipe of radius ‘r’ carries a current. The magnetic flux density out side the pipe at a distance 3r/2 from the axis is 0.04 T. The magnetic flux density inside the pipe at a distance r/2 from the axis will be
(a) 0.01 T (b) 0.02 T (c) 0.04 T (d) zero (e) infinite
Don’t be distracted by the value of the field out side. You should remember that the magnetic flux density at any point inside a long current carrying pipe is zero. The correct option therefore is (d).
(a) 0.01 T (b) 0.02 T (c) 0.04 T (d) zero (e) infinite
Don’t be distracted by the value of the field out side. You should remember that the magnetic flux density at any point inside a long current carrying pipe is zero. The correct option therefore is (d).
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