A solenoid 15cm long has 300 turns of wire. A current of 5A flows through it. What is the magnitude of the magnetic field inside the solenoid?
Correct answer: C. 4πx10-3T
- A. 75 x 10^7T
- B. 60 x 10^4T
- C. 4πx10-3T
- D. 750π x103T
Explanation
The magnetic field inside a solenoid can be calculated using the formula:B = µ0 x n x Iwhereµ0 = permittivity of free space = 4πx10-7I = Current = 5An = is the number of turns per unit length (turns per meter) = 300/0.15 = 2000Substitute these values in the formulaB = 4πx10-7 x 5 x 2000.B = 4πx10-7 x 1x104B = 4πx10-3.
Last updated
About Electromagnetism
Magnetic flux density and magnetic flux describe the strength of a magnetic field and the field passing through a surface. A charged particle moving through a magnetic field experiences a force perpendicular to its velocity and may follow circular or helical motion, depending on the angle between velocity and field.
Practise Electromagnetism
1,176 free Electromagnetism MCQs from Physics, each with the correct answer and an explanation. Unlimited attempts, no account needed.
Exams that ask Physics questions like this
Physics is on 13 papers prepared for on TestUstad, and all of them draw the same bank, so this question is worth knowing for every one of them.
Related questions
1 Tesla is equal to:
1 tesla is equal to:
2.5π µF capacitor and 3000-ohm resistance are joined in series to an AC source of 200 volts and frequency. The power factor of the circuit and the power dissipated in it will respectively be:
A 10 cm long solenoid has 100 turns. What will be the magnetic field inside it along its axis if one-microampere current is passed through it?
A 100-turn coil of area 0.1 m2 rotates at half a revolution per second. It is placed in a uniform magnetic field of 0.01 T perpendicular to the axis of rotation of the coil. Calculate the maximum voltage generated in the coil?