A circular coil 15.0 cm in diameter and containing nine loops lies flat on the g

Question

A circular coil 15.0 cm in diameter and containing nine loops lies flat on the ground. The Earth's magnetic field at this location has magnitude 5.50 Times 10-5 T and points into the Earth at an angle of 53.0 degree below a line pointing due north. A 7.80-A clockwise current passes through the coil. Determine the torque on the coil. Express your answer using three significant figures and include the appropriate units. Which edge of the coil rises up: north, east, south, or west? west east north south

Explanation / Answer

torque for a coil carrying current i and placed in a magnetic field B= × B

where is the N*I*A (current * area) N- no of turns in coil
Therefore, magnetic field= N*I*A*B sin(theta)

In our case,

N=9 , A=Pi * R2 , I=7.8A B= 5.5* 10-5

Therefore Torque= 9*3.14*(0.075)2*7.8*5.5* 10-5 sin(53)

= 3.91* 10-5 Nm

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