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NCERT Class XII Chapter
Magnetism and Matter
Questions With Solutions

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Question : 19 of 25
Marks: +1, -0
A telephone cable at a place has four long straight horizontal wires carrying a current of 1.0 A in the same direction east to west. The earth’s magnetic field at the place is 0.39 G, and the angle of dip is 35°. The magnetic declination is nearly zero. What are the resultant magnetic fields at point 4.0 cm below the cable?
Solution:  
Let us first decide the directions which can best represent the situation.
Here, BHB_H = B cos δ = 0.39 × cos 35° G
BHB_H = 0.32 G
and BVB_V = B sin δ = 0.39 × sin 35° G
BVB_V = 0.22 G
Telephone cable carry a total current of 4.0 A in direction east to west. We want resultant magnetic field 4.0 cm below.
BwireB_{\text{wire}} = (BH−Bwire)2+BV2\sqrt{(B_H-B_{\text{wire}})^2+B_V^2}
BnetB_{\text{net}} = (0.12)2+(0.22)2\sqrt{(0.12)^2+(0.22)^2} = 0.0144+0.0484\sqrt{0.0144+0.0484} = 0.25 G
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