QJEEM14Ph5930: JEE Main 2014 Physics - Class 12 JEEM14Ph5930 Solution: A conductor lies along the -axis at and carries a fixed current …

Question

Verified Solution • 2025–26 Session moderate

A conductor lies along the

z

-axis at

- 1.5 \le z &lt; 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.

Accepted Answer

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}

= 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

Step-by-Step Verified Solution