A circular loop of radius 5 cm carries a current of 0.5A . Calculate the magnitude of the magnetic field at its centre.
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Answered by
35
HEY BUDDY..!!!
HERE'S THE ANSWER..
_____________________
♠️We know magnetic field at centre of a ring is given by
✔️ B = u° × I / 2 × r
⏺️B is magnetic field = ?
⏺️I ( current following in ring ) = 5 A
⏺️r ( radius of ring ) = 5cm = 5 × 10^-2 m
⏺️u° ( magnetic permeability due to free space ) = 4 π × 10^-7 T m / A
♠️ B = u° × I / 2 × r
=> B = ( 4 π × 10^-7 × 5 ) / ( 2 × 10^-2 )
=> B = ( 20 π × 10^-7 ) ( 10 × 10^-2 )
=> B = ( 20 π × 10^-7 × 10 )
=> B = ( 2 π × 10^-5 ) T ✔️✔️
Or
=> B = ( 0.2 π ) G ✔️✔️
✔️As { 1 T = 10^4 G}
✔️G is Gauss
✔️T is Tesla
HOPE HELPED..
JAI HIND..
:-)
HERE'S THE ANSWER..
_____________________
♠️We know magnetic field at centre of a ring is given by
✔️ B = u° × I / 2 × r
⏺️B is magnetic field = ?
⏺️I ( current following in ring ) = 5 A
⏺️r ( radius of ring ) = 5cm = 5 × 10^-2 m
⏺️u° ( magnetic permeability due to free space ) = 4 π × 10^-7 T m / A
♠️ B = u° × I / 2 × r
=> B = ( 4 π × 10^-7 × 5 ) / ( 2 × 10^-2 )
=> B = ( 20 π × 10^-7 ) ( 10 × 10^-2 )
=> B = ( 20 π × 10^-7 × 10 )
=> B = ( 2 π × 10^-5 ) T ✔️✔️
Or
=> B = ( 0.2 π ) G ✔️✔️
✔️As { 1 T = 10^4 G}
✔️G is Gauss
✔️T is Tesla
HOPE HELPED..
JAI HIND..
:-)
riddhi66:
That Jai Hind line in the end ...really inspires..!
Answered by
11
B = nue not × I / 2 r
B = 4π×10^-7 × 5 / 2×0.05
B= 2π × 10^-6/0.1
B= 2π × 10^-5 T
B = 4π×10^-7 × 5 / 2×0.05
B= 2π × 10^-6/0.1
B= 2π × 10^-5 T
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