JEE MAIN - Physics (2014 (Offline) - No. 11)
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 $$\overrightarrow B = 3.0 \times {10^{ - 4}}\,{e^{ - 0.2x}}\,\,{\widehat a_y}\,\,T,$$ find the power required to move the conductor at constant speed to $$x=2.0$$ $$m$$, $$y=0$$ $$m$$ in $$5 \times {10^{ - 3}}s.$$ Assume parallel motion along the $$x$$-axis.
_en_11_1.png)
$$1.57W$$
$$2.97W$$
$$14.85$$ $$W$$
$$29.7W$$
Explanation
Work done in moving the conductor is,
$$W = \int_0^2 {Fdx} $$
$$ = \int_0^2 {3.0 \times {{10}^{ - 4}}\,{e^{ - 02x}} \times 10 \times 3dx} $$
_en_11_2.png)
$$ = 9 \times {10^{ - 3}}\int_0^2 {{e^{ - 0.2x}}\,dx} $$
$$ = {{9 \times {{10}^{ - 3}}} \over {0.2}}\left[ { - {e^{ - 0.2 \times 2}} + 1} \right]\,\,\,$$
$$B = 3.0 \times {10^{ - 4}}\,{e^{ - 0.2x}}$$
(By exponential function)
$$ = {{9 \times {{10}^{ - 3}}} \over {0.2}} \times \left[ {1 - {e^{ - 0.4}}} \right]$$
$$ = 9 \times {10^{ - 3}} \times \left( {0.33} \right) = 2.97 \times {10^{ - 3}}J$$
Power required to move the conductor is,
$$P = {W \over t}$$
$$P = {{2.97 \times {{10}^{ - 3}}} \over {\left( {0.2} \right) \times 5 \times {{10}^{ - 3}}}} = 2.97W$$
$$W = \int_0^2 {Fdx} $$
$$ = \int_0^2 {3.0 \times {{10}^{ - 4}}\,{e^{ - 02x}} \times 10 \times 3dx} $$
$$ = 9 \times {10^{ - 3}}\int_0^2 {{e^{ - 0.2x}}\,dx} $$
$$ = {{9 \times {{10}^{ - 3}}} \over {0.2}}\left[ { - {e^{ - 0.2 \times 2}} + 1} \right]\,\,\,$$
$$B = 3.0 \times {10^{ - 4}}\,{e^{ - 0.2x}}$$
(By exponential function)
$$ = {{9 \times {{10}^{ - 3}}} \over {0.2}} \times \left[ {1 - {e^{ - 0.4}}} \right]$$
$$ = 9 \times {10^{ - 3}} \times \left( {0.33} \right) = 2.97 \times {10^{ - 3}}J$$
Power required to move the conductor is,
$$P = {W \over t}$$
$$P = {{2.97 \times {{10}^{ - 3}}} \over {\left( {0.2} \right) \times 5 \times {{10}^{ - 3}}}} = 2.97W$$
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