21. The force per unit length existing between two infinite parallel conductors is given by ___________
A. µ0i1i2/2πd
B. µ0i1i2/2d
C. µ0i1/2πdi2
D. µ0i1i2/d
Answer: A
Force of the infinite conductor is given by Bil if B, I, and l are mutually perpendicular
F/l = Bi = µ0i1i2/2πd
where d is the separation of two infinite parallel conductors.
22. If the radius of the current-carrying conductor increases, what is the effect on the force?
A. increases
B. decreases
C. remain same
D. become zero
Answer: C
Force of the infinite conductor is given by Bil if B, I, and l are mutually perpendicular
F/l = Bi = µ0i1i2/2πd
where
d is the separation of two infinite parallel conductors. It doesn’t depend on the radius of the conductor so the force remains the same.
23. If the current of the conductor increases, what is the effect on the force?
A. increases
B. decreases
C. remain same
D. become zero
Answer: A
The Force of the infinite conductor is given by Bil
where
B, I, and l are mutually perpendicular
F/l = Bi = µ0i1i2/2πd
where
d is the separation of two infinite parallel conductors. So, as current increases force increases.
24. If the length of the current-carrying conductor increases, what is the effect on the force?
A. increases
B. decreases
C. remain same
D. become zero
Answer: A
Force of the infinite conductor is given by Bil where
B, I, and l are mutually perpendicular
F/l = Bi = µ0i1i2/2πd
Where
d is a separation of two infinite parallel conductors. So, as the length of the conductor increases force increases.
25. The magnetic field strength of a solenoid can be increased by inserting which of the following materials as the core?
A. Copper
B. Silver
C. Iron
D. Aluminium
Answer: C
The Magnetic field of a solenoid increases when we insert an iron core because iron is a ferromagnetic material and ferromagnetic materials help in increasing the magnetic property.
26. If a coil is wound around a steel core and electric current is passed through the coil, the steel core acts as a?
A. Electromagnet
B. Permanent magnet
C. Neither electromagnet nor permanent magnet
D. Either electromagnet or permanent magnet
Answer: B
When a coil is wound around a steel core, the steel core behaves like a permanent magnet because it is a ferromagnetic material and once it becomes magnetic it does not lose its magnetic property.
27. The current in a solenoid is 30A, and the number of turns per unit length is 500 turns per meter. Calculate the magnetic field if the core is air.
A. 18.84T
B. 18.84mT
C. 1.84T
D. 1.84mT
Answer: B
The magnetic field in a solenoid is given by:
B = μnI
Substituting the values in the given values in the equation,
B = 18.84mT.
28. The magnetic field of the solenoid is 18.84mT, and the number of turns per unit length is 500 turns per meter. Calculate the current if the core is air.
A. 300A
B. 30A
C. 3A
D. 300mA
Answer: B
The magnetic field in a solenoid is given by:
B = μnI
Substituting the values in the given values in the equation, I = 30A.
29. The magnetic field of the solenoid is 18.84mT, and the current is 30A. Calculate the number of turns per unit length if the core is air.
A. 1500 turns/m
B. 1000 turns/m
C. 500 turns /m
D. 2000 turns/m
Answer: C
The magnetic field in a solenoid is given by:
B = μnI
Substituting the values in the given values in the equation n = 500 turns/m.
