A metal rod of length L rotates about one end at origin with a uniform angular velocity ω. The magnetic field radially falls off as B(r) = B₀ e⁻λr; λ being a positive constant. The emf induced (neglecting the centripetal force on electrons in the rod) is :

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q36 – JEE-MAIN 2026

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Home Physics EM InductionQuestion #146

A metal rod of length

L

rotates about one end at origin with a uniform angular velocity

ω

. The magnetic field radially falls off as

B (r) = B_{0} e^{- λ r}

;

λ

being a positive constant. The emf induced (neglecting the centripetal force on electrons in the rod) is :

Clue 1 of 3: Motional EMF

The EMF induced in a conductor moving in a magnetic field is due to the Lorentz force on the charge carriers.

🧠 Full Solution Path

Step 1: Determine the differential EMF✦ Active

Consider a small segment $d r$ of the rod at a distance $r$ from the origin. The velocity of this segment is $v = ω r$ . The magnetic field at this point is $B (r) = B_{0} e^{- λ r}$ . Since the rod rotates and the magnetic field is radial, the velocity $\vec{v}$ is tangential and perpendicular to the magnetic field $\vec{B}$ . The induced EMF across this segment $d r$ is:

d E = B (r) v d r = (B_{0} e^{- λ r}) (ω r) d r = B_{0} ω r e^{- λ r} d r

Step 2: Set up the integral for total EMF○ Expand

To find the total EMF induced across the entire rod of length $L$ , we integrate $d E$ from $r = 0$ to $r = L$ :

E = \int_{0}^{L} B_{0} ω r e^{- λ r} d r = B_{0} ω \int_{0}^{L} r e^{- λ r} d r

Step 3: Evaluate the integral○ Expand

We use integration by parts for $\int r e^{- λ r} d r$ . Let $u = r$ and $d v = e^{- λ r} d r$ . Then $d u = d r$ and $v = - \frac{1}{λ} e^{- λ r}$ . The integral becomes:

\int r e^{- λ r} d r = r (- \frac{1}{λ} e^{- λ r}) - \int (- \frac{1}{λ} e^{- λ r}) d r = - \frac{r}{λ} e^{- λ r} - \frac{1}{λ^{2}} e^{- λ r} = - e^{- λ r} (\frac{r}{λ} + \frac{1}{λ^{2}})

Now, evaluate the definite integral from $0$ to $L$ :

E = B_{0} ω {[- e^{- λ r} (\frac{r}{λ} + \frac{1}{λ^{2}})]}_{0}^{L}

Substitute the limits:

E = B_{0} ω [(- e^{- λ L} (\frac{L}{λ} + \frac{1}{λ^{2}})) - (- e^{0} (\frac{0}{λ} + \frac{1}{λ^{2}}))]

E = B_{0} ω [- e^{- λ L} (\frac{L}{λ} + \frac{1}{λ^{2}}) + \frac{1}{λ^{2}}]

E = B_{0} ω [\frac{1}{λ^{2}} - e^{- λ L} (\frac{1}{λ^{2}} + \frac{L}{λ})]

This result matches option 1.

💡 Teacher's Secret Hint

Be careful with the signs and the evaluation of the limits, especially the term at $r = 0$ where $e^{0} = 1$ .

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Physics Question 36 – JEE-MAIN 2026

Clue 1 of 3: Motional EMF

🧠 Full Solution Path