Question 227: Magnetism and Matter

Question

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

Accepted Answer

Work done in moving the conductor is,
$$W = \int_0^2 {Fdx} $$
$$ = \int_0^2 {3.0 	imes {{10}^{ - 4}}\,{e^{ - 02x}} 	imes 10 	imes 3dx} $$
$$ = 9 	imes {10^{ - 3}}\int_0^2 {{e^{ - 0.2x}}\,dx} $$
$$ = {{9 	imes {{10}^{ - 3}}} \over {0.2}}\left[ { - {e^{ - 0.2 	imes 2}} + 1} ight]\,\,\,$$
$$B = 3.0 	imes {10^{ - 4}}\,{e^{ - 0.2x}}$$
(By exponential function)
$$ = {{9 	imes {{10}^{ - 3}}} \over {0.2}} 	imes \left[ {1 - {e^{ - 0.4}}} ight]$$
$$ = 9 	imes {10^{ - 3}} 	imes \left( {0.33} ight) = 2.97 	imes {10^{ - 3}}J$$
Power required to move the conductor is,
$$P = {W \over t}$$
$$P = {{2.97 	imes {{10}^{ - 3}}} \over {\left( {0.2} ight) 	imes 5 	imes {{10}^{ - 3}}}} = 2.97W$$

Answer

$$1.57W$$

Answer

$$2.97W$$

Answer

$$14.85$$  $$W$$

Answer

$$29.7W$$

Magnetism and Matter

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