A rectangular loop PQRS is formed by joining the corners of cube of side 20 cm as shown in figure. If I = 8 A and B = 0.4 i ^ T then the magnitude of torque on loop is P × 1 0 − 3 N-m. Find the value of P.

M = I ( PQ × QR ) = 8 [ 0.2 k ^ × ( 0.2 i ^ − 0.2 j ^ )] M = 0.32 ( i ^ + j ^ ) A m 2 τ = M × B = 0.32 ( i ^ + j ^ ) × 0.4 i ^ = − 0.128 k ^ N m ∣ τ ∣ = 0.128 = 128 × 1 0 − 3 N m P = 128

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PhysicsMagnetismMagnetic torque & energyMedium2 minQB

PhysicsMediuminteger

A rectangular loop PQRS is formed by joining the corners of cube of side 20 cm as shown in figure. If $I = 8$ A and $B = 0.4 \hat{i}$ T then the magnitude of torque on loop is $P \times 1 0^{- 3}$ N-m. Find the value of P.

Answer:

128

Solution:

$M = I (PQ \times QR) = 8 [0.2 \hat{k} \times (0.2 \hat{i} - 0.2 \hat{j})]$

$M = 0.32 (\hat{i} + \hat{j}) A m^{2}$

$τ = M \times B = 0.32 (\hat{i} + \hat{j}) \times 0.4 \hat{i} = - 0.128 \hat{k} N m$

$∣ τ ∣ = 0.128 = 128 \times 1 0^{- 3} N m$

$P = 128$

Stream:JEESubject:PhysicsTopic:MagnetismSubtopic:Magnetic torque & energy

⏱ 2mℹ️ Source: QB

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