Question

if a conductor of length 1m is moved with velocity v across a magnetic field B at an angle of 30 °with B, then the motional emf will be
(A) vBl
(B) 1÷2vBl
(C) 1÷2vB
(D) 0.866vB​

Answer 1

Explanation:

magnetic force on length = I L B sin30⁰ =( I L B) /2

work done = F x ds =(I L B) /2 x ds

emf = work done /dq = [( I L B) x ds] /2 dq

as I = dq/dt

emf = ( L B x ds/dt) /2 = V L B /2

as L = 1m

emf = VB/2

Answer 2

Concept:

The motional emf is represented by the equation- emf, E = Blvsinθ.

Given:

Length of a conductor = 1m

The velocity of a conductor = v

Magnetic field = B

Angle = 30°

Find:

We need to determine the motional emf of a conductor.

Solution:

An emf is produced when a coil is moved near a magnet, and the opposite is true when a magnet is moved toward a coil. Motion produces what is loosely referred to as motional emf in a magnetic field that is stationary with respect to the Earth.

The Motional EMF is expressed as- emf, E = Bℓvsinθ

where B is the magnetic field,

ℓ is the length of the conductor,

v is the velocity while E is the motional emf.

We have length, l = 1m, velocity = v and Magnetic field as B having θ  as 30°.

Therefore, the motional emf equation becomes, E = Bvsin30°

E = 1/2vB

Thus,  the motional emf will be 1/2vB.

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