A wire PQRST carrying current I =5A is placed in uniform magnetic field B=2T as shown in figure if thr length of the part QR=4cm and SR=6cm then the magnetic force onSR edge of the wire is
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Answer:
Given that,
SR = 6 cm
QR = 4 cm
Magnetic force on arm SR is F=BILsinθ
Where, B is magnetic field
From the figure, sinθ=
6
4
=
3
2
So, F=2×5×0.06×
3
2
F=0.4N
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