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# Test: Magnetic Force And Lorentz Force

## 10 Questions MCQ Test Electromagnetic Theory | Test: Magnetic Force And Lorentz Force

Description
This mock test of Test: Magnetic Force And Lorentz Force for Electrical Engineering (EE) helps you for every Electrical Engineering (EE) entrance exam. This contains 10 Multiple Choice Questions for Electrical Engineering (EE) Test: Magnetic Force And Lorentz Force (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Magnetic Force And Lorentz Force quiz give you a good mix of easy questions and tough questions. Electrical Engineering (EE) students definitely take this Test: Magnetic Force And Lorentz Force exercise for a better result in the exam. You can find other Test: Magnetic Force And Lorentz Force extra questions, long questions & short questions for Electrical Engineering (EE) on EduRev as well by searching above.
QUESTION: 1

### Find the electric force when the charge of 2C is subjected to an electric field of 6 units.

Solution:

Explanation: The electric force is given by F = qE, where q = 2C and E = 6 units. Thus we get F = 2 x 6 = 12 units.

QUESTION: 2

### Find the magnetic force when a charge 3.5C with flux density of 4 units is having a velocity of 2m/s.

Solution:

Explanation: The magnetic force is given by F = q(v x B), where q = 3.5C, v = 2m/s and B = 4 units. Thus we get F = 3.5(2 x 4) = 28 units.

QUESTION: 3

### Find the electric field when the velocity of the field is 12m/s and the flux density is 8.75 units.

Solution:

Explanation: The electric field intensity is the product of the velocity and the magnetic flux density ie, E = v x B = 12 x 8.75 = 105 units.

QUESTION: 4

Find the Lorentz force of a charge 2.5C having an electric field of 5 units and magnetic field of 7.25 units with a velocity 1.5m/s.

Solution:

Explanation: The Lorentz force is given by F = qE + q(v x B), it is the sum of electric and magnetic force. On substituting q = 2.5, E = 5, v = 1.5 and B = 7.25, F = 2.5(5) + 2.5(1.5 x 7.25) = 39.68 units.

QUESTION: 5

The force on a conductor of length 12cm having current 8A and flux density 3.75 units at an angle of 300 is

Solution:

Explanation: The force on a conductor is given by F = BIL sin θ, where B = 3.75, I = 8, L = 0.12 and θ = 300. We get F = 3.75 x 8 x 0.12 sin 30 = 1.8 units.

QUESTION: 6

The force per unit length of two conductors carrying equal currents of 5A separated by a distance of 20cm in air(in 10-6 order)

Solution:

Explanation: The force per unit length of two conductors is given by
F = μ I1xI2/2πD, where I1 = I2 = 5 and D = 0.2. Thus F = 4π x 10-7 x 52/ 2π x 0.2 = 25 x 10-6 units.

QUESTION: 7

When currents are moving in the same direction in two conductors, then the force will be

Solution:

Explanation: When two conductors are having currents moving in the same direction then the forces of the two conductors will be moving towards each other or attractive.

QUESTION: 8

Find the flux density due to a conductor of length 6m and carrying a current of 3A(in 10-7order)

Solution:

Explanation: The flux density is B = μH, where H = I/2πR. Put I = 3 and R = 6, we get B = 4π x 10-7 x 3/2π x 6 = 1 x 10-7 units.

QUESTION: 9

Find the maximum force of the conductor having length 60cm, current 2.75A and flux density of 9 units.

Solution: