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Physics Question on Moving charges and magnetism

A conductor lies along the zz -axis at −1.5≤z<1.5 m-1.5 \leq z<1.5 \,m and carries a fixed current of 10.0A10.0 A in −a^z-\hat{a}_{z} direction (see figure). For a field B⃗=3.0×10−4e−0.2xa^yT\vec{B}=3.0 \times 10^{-4} e^{-0.2 x} \hat{a}_{y} T, find the power required to move the conductor at constant speed to x=2.0m,y=0 mx=2.0 m , y=0 \,m in 5×10−3 s5 \times 10^{-3} \,s. Assume parallel motion along the xx -axis

A

1.57 W

B

2.97 W

C

14.85 W

D

29.7 W

Answer

2.97 W

Explanation

Solution

Average Power = work  time =\frac{\text { work }}{\text { time }}
W=∫02FdxW =\int_{0}^{2} F dx
=∫023.0×10−4e−0.2x×10×3 dx=\int_{0}^{2} 3.0 \times 10^{-4} e^{-0.2 x} \times 10 \times 3 \,d x
=9×10−3∫02e−0.2x dx=9 \times 10^{-3} \int_{0}^{2} e^{-0.2 x} \,dx
=9×10−30.2[−e−0.2×2+1]=\frac{9 \times 10^{-3}}{0.2}\left[-e^{-0.2 \times 2}+1\right]
B=3.0×10−4e−0.2xB=3.0 \times 10^{-4} e^{-0.2 x}

=9×10−30.2×[1−e−0.4]=\frac{9 \times 10^{-3}}{0.2} \times\left[1-e^{-0.4}\right]
=9×10−3×(0.33)=9 \times 10^{-3} \times(0.33)
=2.97×10−3 J=2.97 \times 10^{-3} \,J
P=2.97×10−3(0.2)×5×10−3P=\frac{2.97 \times 10^{-3}}{(0.2) \times 5 \times 10^{-3}}
=2.97 W=2.97 \,W