A conductor lies along the z-axis at $-1.5 \leq \mathrm{z}<1.5 \mathrm{~m}$ and carries a fixed current of 10.0 A in $-\hat{\mathrm{a}}_{\mathrm{z}}$ direction (see figure). For a field $\overrightarrow{\mathrm{B}}=3.0 \times 10^{-4} \mathrm{e}^{-0.2 \mathrm{x}} \hat{\mathrm{a}}_{\mathrm{y}} \mathrm{T}$, find the power required to move the conductor at constant speed to $\mathrm{x}=2.0 \mathrm{~m}, \mathrm{y}=0 \mathrm{~m}$ in $5 \times 10^{-3} \mathrm{~s}$. Assume parallel motion along the x-axis

JEE Main · 2014 · 06 Apr 2014 · Offline · Electromagnetic Induction

A conductor lies along the z-axis at

- 1.5 \leq z < 1.5 m

and carries a fixed current of 10.0 A in

- \overset{a}{^}_{z}

direction (see figure). For a field

B = 3.0 \times 1 0^{- 4} e^{- 0.2 x} \overset{a}{^}_{y} T

, find the power required to move the conductor at constant speed to

x = 2.0 m, y = 0 m

5 \times 1 0^{- 3} s

. Assume parallel motion along the x-axis

Answer: (D)

Power is the work done against the magnetic force divided by the time taken. The force on the current-carrying length

ℓ = 3

m is

F = I ℓ B

, and since

B = 3 \times 1 0^{- 4} e^{- 0.2 x}

falls off with position, the work from

x = 0

x = 2

m is

W = \int_{0}^{2} I ℓ B d x = (10) (3) (3 \times 1 0^{- 4}) \int_{0}^{2} e^{- 0.2 x} d x

. That integral is

[- 5 e^{- 0.2 x}]_{0}^{2} = 5 (1 - e^{- 0.4})

, so

W = 9 \times 1 0^{- 3} \times 5 (1 - e^{- 0.4}) \approx 1.49 \times 1 0^{- 2}

J. Dividing by the time

5 \times 1 0^{- 3}

s leaves

P = 9 (1 - e^{- 0.4}) \approx 2.97

JEE Main · 2014 · 06 Apr 2014 · Offline · Single correct option