JEE Challenger
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PhyHardMain 2026 • 5th Apr - Shift 2

Induced EMF in Rotating Metal Rod in Radial Exponential Magnetic Field

A metal rod of length LL rotates about one end at origin with a uniform angular velocity ω\omega. The magnetic field radially falls off as B(r)=B0e−λrB(r) = B_0 e^{-\lambda r}; λ\lambda being a positive constant. The emf induced (neglecting the centripetal force on electrons in the rod) is :

Options

A

B0ω[1λ2−e−λL(1λ2+Lλ)]B_0 \omega \left[ \frac{1}{\lambda^2} - e^{-\lambda L} \left(\frac{1}{\lambda^2} + \frac{L}{\lambda}\right) \right]

Correct
B

B0ω[1λ2+e−λL(1λ2+Lλ)]B_0 \omega \left[ \frac{1}{\lambda^2} + e^{-\lambda L} \left(\frac{1}{\lambda^2} + \frac{L}{\lambda}\right) \right]

C

B0ω[4λ2−e−2λL(1λ2+2Lλ)]B_0 \omega \left[ \frac{4}{\lambda^2} - e^{-2\lambda L} \left(\frac{1}{\lambda^2} + \frac{2L}{\lambda}\right) \right]

D

B0ω[3λ2−e−3λL(3λ2+Lλ)]B_0 \omega \left[ \frac{3}{\lambda^2} - e^{-3\lambda L} \left(\frac{3}{\lambda^2} + \frac{L}{\lambda}\right) \right]

Topics & Concepts

Electromagnetic InductionFaraday's Law

Step-by-Step Solution

To find the total induced electromotive force (emf) in the rotating metal rod, we calculate the motional emf induced across a small element of the rod and integrate it over its entire length.

Step 1: Induced EMF in a Small Element

Consider a small element of length drdr located at a distance rr from the origin (the axis of rotation).

Since the rod rotates with a uniform angular velocity ω\omega, the linear velocity of this element is given by: v(r)=ωrv(r) = \omega r

The magnetic field perpendicular to the direction of motion at a distance rr is: B(r)=B0e−λrB(r) = B_0 e^{-\lambda r}

The motional emf dEd\mathcal{E} induced in this small element drdr is: dE=v(r)B(r)dr=(ωr)(B0e−λr)dr=B0ωre−λrdrd\mathcal{E} = v(r) B(r) dr = (\omega r)\left(B_0 e^{-\lambda r}\right) dr = B_0 \omega r e^{-\lambda r} dr

Step 2: Integrating Over the Entire Length

The total emf E\mathcal{E} induced across the entire rod of length LL is obtained by integrating dEd\mathcal{E} from r=0r = 0 to r=Lr = L: E=∫0LB0ωre−λrdr=B0ω∫0Lre−λrdr\mathcal{E} = \int_0^L B_0 \omega r e^{-\lambda r} dr = B_0 \omega \int_0^L r e^{-\lambda r} dr

Step 3: Evaluating the Integral

We integrate using the method of integration by parts, where ∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du:

Let:

  • u=r  ⟹  du=dru = r \implies du = dr
  • dv=e−λrdr  ⟹  v=−1λe−λrdv = e^{-\lambda r} dr \implies v = -\frac{1}{\lambda} e^{-\lambda r}

Applying integration by parts: ∫re−λrdr=−rλe−λr−∫(−1λe−λr)dr\int r e^{-\lambda r} dr = -\frac{r}{\lambda} e^{-\lambda r} - \int \left( -\frac{1}{\lambda} e^{-\lambda r} \right) dr ∫re−λrdr=−rλe−λr−1λ2e−λr=−e−λr(rλ+1λ2)\int r e^{-\lambda r} dr = -\frac{r}{\lambda} e^{-\lambda r} - \frac{1}{\lambda^2} e^{-\lambda r} = -e^{-\lambda r} \left( \frac{r}{\lambda} + \frac{1}{\lambda^2} \right)

Now, evaluating from the limits 00 to LL: ∫0Lre−λrdr=[−e−λr(rλ+1λ2)]0L\int_0^L r e^{-\lambda r} dr = \left[ -e^{-\lambda r} \left( \frac{r}{\lambda} + \frac{1}{\lambda^2} \right) \right]_0^L =(−e−λL(Lλ+1λ2))−(−e0(0+1λ2))= \left( -e^{-\lambda L} \left( \frac{L}{\lambda} + \frac{1}{\lambda^2} \right) \right) - \left( -e^{0} \left( 0 + \frac{1}{\lambda^2} \right) \right) =1λ2−e−λL(1λ2+Lλ)= \frac{1}{\lambda^2} - e^{-\lambda L} \left( \frac{1}{\lambda^2} + \frac{L}{\lambda} \right)

Step 4: Final Expression

Substituting this result back into the emf equation gives: E=B0ω[1λ2−e−λL(1λ2+Lλ)]\mathcal{E} = B_0 \omega \left[ \frac{1}{\lambda^2} - e^{-\lambda L} \left( \frac{1}{\lambda^2} + \frac{L}{\lambda} \right) \right]

Thus, the correct option is A.

Induced EMF in Rotating Metal Rod in Radial Exponential Magnetic Field | Physics PYQ Solution - JEE Challenger