250+ TOP MCQs on Magnetic Field on the Axis of a Circular Current Loop | Class12 Physics

Physics Written Test Questions for Schools on “Magnetic Field on the Axis of a Circular Current Loop”.

1. Pick out the expression for magnetic field strength at any point at the center of a circular loop from the following?
a) B = ( [ frac {mu_o I}{4 pi r} ] int_0^{2pi r} )dl sin⁡90
b) B = ( [ frac {mu_o I}{4 pi r} ] int_0^{2pi r} )dl sin⁡45
c) B = ( [ frac {mu_o I}{4 pi r} ] int_0^{2pi r} )dl sin⁡30
d) B = [μ_o × 4πr]( int_0^{2pi r} )dl sin⁡90
Answer: a
Clarification: The magnetic field strength at any point at center of circular loop carrying current I and radius r is given as:
B = ( [ frac {mu_o I}{4 pi r} ] int_0^{2pi r} )dl sin⁡90
It can also be expressed as → B = ( [ frac {mu_o I}{4 pi r} ] ) × 2πr or B = ( [ frac {mu_o I}{2 r} ] ) direction.
It is inwards if the current is flowing in the clockwise direction and it is onwards if the current is flowing in the anticlockwise direction.

2. Identify the expression for the magnetic field on the axis of circular loop.
a) B=( frac {mu_o Ir}{4 pi big ( (r^2 , + , x^2)^{frac {1}{2}} big ] } ) × 2πr
b) B=( frac {mu_o Ir}{4 pi big ( (r^2 , + , x^2)^{frac {3}{2}} big ] } )
c) B=( frac {mu_o Ir}{4 pi big ( (r^2 , + , x^2)^{frac {3}{2}} big ] } ) × 2πr
d) B=( frac {mu_o Ir}{4 pi big ( (r^2 , + , x^2)^{frac {1}{2}} big ] } ) × 4πr
Answer: c
Clarification: The expression for magnetic field on the axis of circular loop is given as:
B=( frac {mu_o Ir}{4 pi big ( (r^2 , + , x^2)^{frac {3}{2}} big ] } ) × 2πr
It is towards the loop if current in it is in clockwise direction and it is away from the loop if current in it is in anticlockwise direction.

3. The magnetic field due to a current carrying circular loop of radius 4 cm at a point on the axis at a distance of 7 cm from the center is 48 μT. What will be the value at the center of the loop?
a) 390 μT
b) 393 μT
c) 395 μT
d) 397 μT
Answer: b
Clarification: Field along axis of coil → B =( frac {mu_o IR^2}{2 big ( (R^2 , + , x^2)^{frac {3}{2}} big ] } )
At the coil of the coil → B₁ = ( frac {mu_o I}{2R})
(frac {B_1}{B} = frac {mu_o I}{2R} , times , frac {2 big ( (R^2 , + , x^2)^{frac {3}{2}} big ] }{mu_o IR^2} = frac {2 big ( (R^2 , + , x^2)^{frac {3}{2}} big ] }{R^3})
B1 = (frac {48 [4^2 + 7^2]^{frac {3}{2}}}{4^3})
B1 = (frac {48 times (65)^{frac {3}{2}}}{4^3})
B1 = 131.01 × 3
B1 = 393.03 = 393 μT

4. When an arc of a circle of radius R subtends an angle of (frac {pi }{4}) at the center, and carries a current I, the magnetic field at the center is B = (frac {mu_o I}{16R}).
a) True
b) False
Answer: a
Clarification: Magnetic field at the center of a circular arc of radius R, carries a current I and makes an angle θ at the center is given by
B = (frac {mu_o I theta}{4 pi R})
In this case θ = (frac {pi }{4})
B = (frac {mu_o I (frac {pi }{4})}{4 pi R})
B = (frac {mu_o I}{16 R})

5. A horizontal overhead power line carries a current of 100A in east to west direction. What is the magnitude and direction of the magnetic field due to the current, 2 m below the line?
a) 3 x 10^-5 T
b) 2 x 10^-5 T
c) 1 x 10^-5 T
d) 4 x 10^-5 T
Answer: c
Clarification: Given: I = 100A; r = 2 m; (frac {mu_o}{4 pi }) = 10^-7
The required equation → B = ((frac {mu_o}{4 pi }) , times , frac {2I}{r})
B = 10^-7 × 2 × (frac {100}{2})
B = 10^-7 × 100
B = 1 × 10^-5 T
Therefore, the magnitude of the magnetic field is 1 × 10^-5 T and the direction of the magnetic field is south.

Physics Written Test Questions for Schools,