A metal rod of length rotates about one end at origin with a uniform angular velocity . The magnetic field radially falls

Question

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 [2026]

Smart Achievers · Accepted Answer

(1)

$ε = \int_{0}^{L} B (ω r) d r = B_{0} ω \int_{0}^{L} e^{- λ r} \cdot r d r$

$= B_{0} ω {[r \int e^{- λ r} d r - \int 1 \cdot \frac{e^{- λ r}}{- λ} d r]}_{0}^{L}$

$ε = B_{0} ω {[\frac{r e^{- λ r}}{- λ} - \frac{e^{- λ r}}{λ^{2}}]}_{0}^{L}$

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

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

Solution

Q.
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 [2026]

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

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

3 $B_{0} ω [\frac{4}{λ^{2}} - e^{- 2 λ L} (\frac{1}{λ^{2}} + \frac{2 L}{λ})]$

4 $B_{0} ω [\frac{3}{λ^{2}} - e^{- 3 λ L} (\frac{3}{λ^{2}} + \frac{L}{λ})]$

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Solution

Q. 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)=B0e-λr, λ being a positive constant. The emf induced (neglecting the centripetal force on electrons in the rod) is [2026]

1 B0ω[1λ2-e-λL(1λ2+Lλ)]

2 B0ω[1λ2+e-λL(1λ2+Lλ)]

3 B0ω[4λ2-e-2λL(1λ2+2Lλ)]

4 B0ω[3λ2-e-3λL(3λ2+Lλ)]