Magnetic field due to currentHard
Question
A stationary, circular wall clock has a face with a radius of 15 cm. Six turns of wire are wounded around its perimeter, the wire carries a current 2.0 A in the clockwise direction. The clock is located, where there is a constant, uniform external magnetic field of 70 mT (but the clock still keeps perfect time) at exactly 1 : 00 pm, the hour hand of the clock points in the direction of the external magnetic field
Options
A.Magnitude of the torque on the winding due to the magnetic field is 3.12 × 10-2 N-m.
B.After 20 minutes the minute hand will point in the direction of the torque on the winding due to the magnetic field.
C.Magnitude of the torque on the winding due to the magnetic field is 5.94 × 10-2 N-m.
D.After 30 minutes the minute hand will point in the direction of the torque on the winding due to the magnetic field.
Solution
Torque is directed at right angle to hour hand. hence minute hand will be in the direction of T = M × B × sin θ = NIAB sin 90
=
= 0.0594 NmCreate a free account to view solution
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