SATHEE: Electromagnetic Induction and Alternating Current 7 Question 6

Electromagnetic Induction and Alternating Current 7 Question 6

6. 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 $- a_{z}$ direction (see figure). For a field $B = 3.0 \times 10^{- 4} e^{- 0.2 x} a_{y} T$ , find the power required to move the conductor at constant speed to $x = 2.0 m, y = 0$ in $5 \times 10^{- 3} s$ .

Assume parallel motion along the $x$ -axis.

(2014 Main)

(a) $1.57 W$

(b) $2.97 W$

(d) $29.7 W$

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Answer:

Correct Answer: 6. (b)

Solution:

When force exerted on a current carrying conductor

$F_{e x t} = B I L$

Average power $= \frac{Work done}{Time taken}$

$\begin{aligned} P & = \frac{1}{t} \int_{0}^{2} F_{ext.} \cdot d x = \frac{1}{t} \int_{0}^{2} B (x) I L d x \\ = \frac{1}{5 \times 10^{- 3}} \int_{0}^{2} 3 \times 10^{- 4} e^{- 0.2 x} \times 10 \times 3 d x \\ = 9 [1 - e^{- 0.4}] = 91 - \frac{1}{e^{0.4}} \\ = 2.967 \approx 2.97 W \end{aligned}$

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Electromagnetic Induction and Alternating Current 7 Question 6

6. A conductor lies along the z-axis at −1.5≤z<1.5m and carries a fixed current of 10.0 A in −az direction (see figure). For a field B=3.0×10−4e−0.2xayT, find the power required to move the conductor at constant speed to x=2.0m,y=0 in 5×10−3s.

Answer:

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