A solenoid having length 1.3 m and 2 cm in diameter has 5 turns in each 1 cm of length. A current of 10 A is flowing through it then magnetic field at one end of solenoid on the axis is
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A solenoid having length 1.3 m and 2 cm diameter has 5 turns in each 1 cm of length. A current of 10A is flowing through it.
To find : The magnetic field at one end of solenoid on the axis.
solution : number of turns per cm = 5
so, number of turns in 1.3 m, N = 5 × 130 = 650 turns
current, i = 10 A
now magnetic field at one end on the axis of solenoid is given by, B = μ₀Ni/2
= (4π × 10^-7 × 650 × 10)/2
= 2π × 6500 × 10^-7 T
= 40820 × 10^-7 T
= 4.082 × 10¯³ T
Therefore the magnetic field at one end on the axis of solenoid is 4.082 × 10¯³ T
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