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Magnetics question

2014 · Shift 0 · Q55
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Magnetics question

2014 · Shift 0 · Q55

JEE MainPhysicsMagneticsMCQ+4 / −1
A conductor lies along the zzz-axis at −1.5≤z<1.5 m- 1.5 \le z \lt 1.5\,m−1.5≤z<1.5m and carries a fixed current of 10.0A10.0A10.0A in −a^z- {\widehat a_z}−az​ direction (see figure). For a field B→=3.0×10−4 e−0.2x  a^y  T,\overrightarrow B = 3.0 \times {10^{ - 4}}\,{e^{ - 0.2x}}\,\,{\widehat a_y}\,\,T,B=3.0×10−4e−0.2xay​T, find the power required to move the conductor at constant speed to x=2.0mx=2.0mx=2.0m, y=0my=0my=0m in 5×10−3s.5 \times {10^{ - 3}}s.5×10−3s. Assume parallel motion along the xxx-axis. JEE Main 2014 (Offline) Physics - Magnetic Effect of Current Question 191 English
  1. A
    1.57W1.57W1.57W
  2. B
    2.97W2.97W2.97W
  3. C
    14.85W14.85W14.85W
  4. D
    29.7W29.7W29.7W
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