a solenoid coil of 300 turns per metre is carrying current of 5 A .The lenght of the solenoid is 0.5 m and radius 1 cm.find the magnitude of the magnetic field inside the solenoid
Answers
Answered by
34
sorry there is an error in the pic... i have written 5 in place of 0.5. So the answer would be 1.18 * 10^-4.
here is ur answer. plz mark it as brainliest
here is ur answer. plz mark it as brainliest
Attachments:
Answered by
25
Hope it helps...!!!
-------------------------
The magnitude of magnetic field inside the solenoid is given as,
B=μo nI
where,
μo=4π×10^−7 Tesla/Amp.m
n=number of turns per unit length
I=current
On substituting the values we get,
B=(4π×10^−7 Tesla/Amp.m)×(300 turns/m)×(0.5 Amp)
B=1.88×10^−4 Tesla
--------------------------
#Be Brainly✌️
-------------------------
The magnitude of magnetic field inside the solenoid is given as,
B=μo nI
where,
μo=4π×10^−7 Tesla/Amp.m
n=number of turns per unit length
I=current
On substituting the values we get,
B=(4π×10^−7 Tesla/Amp.m)×(300 turns/m)×(0.5 Amp)
B=1.88×10^−4 Tesla
--------------------------
#Be Brainly✌️
Similar questions