MHT CET · Physics · Electromagnetic Induction

A straight conductor of length $0.4 m$ is moving with a speed of $7 {ms}^{- 1}$ perpendicular to the magnetic field of intensity of $0.9 Wb m^{- 2}$ . The induced emf across the conductor will be

A $7.25 V$
B $5.52 V$
C $1.25 V$
D $2.52 V$

Verified Solution

Answer & Solution

Correct Answer

(D) $2.52 V$

Step-by-step Solution

Detailed explanation

If a rod of length

l

is moved with velocity

\vec{v}

at an angle

θ

to the length of the rod in a filed

\vec{B}

which is perpendicular to the plane of motion, the flux linked with the area generated by the motion of rod in time t.

ϕ = B l (v s i n θ) t

So,

e = \frac{d ϕ}{d t} = B v l s i n θ

This will be maximum when

s i n θ = m a x = 1

i.e., the rod is moving perpendicular to its length and then

{(e)}_{m a x} = B v l

∴ e = 0.9 \times 7 \times 0.4 = 2.52 V

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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A straight conductor of length 0.4 m is moving with a speed of 7 ms-1 perpendicular to the magnetic field of intensity of 0.9 Wb m-2. The induced emf across the conductor will be

Step-by-step Solution

A straight conductor of length $0.4 m$ is moving with a speed of $7 {ms}^{- 1}$ perpendicular to the magnetic field of intensity of $0.9 Wb m^{- 2}$ . The induced emf across the conductor will be