A rectangular loop of sides 20 cm and 10 cm carries a current of 5.0 A. A uniform magnetic field of magnituded 0.20 T exists parallel ot the longer side of the loop (a) What is the force acting on the loop? (b) what is the torque acting on the loop?
Answers
l = 20cm = 20 × 10–2m B = 10cm = 10 × 10–2m i = 5A, B = 0.2T (a) There is no force on the sides AB and CD. But the force on the sides AD and BC are opposite. So they cancel each other. (b) Torque on the loop τ = ni vector A x vector B = niABSin90° = 1 × 5 × 20 × 10–2 × 10× 10–2 0.2 = 2 × 10–2 = 0.02N-M Parallel to the shorter side.
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ANSWER
Given that,
Area A=200 cm2
Current I=5.0 A
Magnetic fieldB=0.20 T
For Loop ABCD
Let AB = 10 cm = CD and BC = DA = 20cm
(a). magnetic moment
M=IA
M=5×200×10−4
M=1000×10−4
M=0.1
Now, force of AB
FAB=I(AB×B)
FAB=5×0.1×0.2
FAB=0.1N
Now, force of CD
FCD=−I(CD×B)
FCD=−5×0.1×0.2
FCD=−0.1N
Now, force of BC
FBC=I(BC×B)
FBC=0
Similarly
FDA=0
Now, the force acting on the loop
Floop=0.1−0.1+0+0
Floop=0
(b).The torque on the loop
τ=M×B
τ=0.1×0.2
τ=0.02Nm
Hence, the force on loop is 0 and the torque on the loop is 0.02 Nm
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