Torque from magnetic field

Electricity and Magnetism Level 2

A rectangular conducting wire $PQRS$ with current intensity $I$ is fixed on the $xy$ -plane where a uniform magnetic field with magnetic intensity $B$ exists, as above. The direction of the magnetic field is $+x$ direction as indicated by the green arrow. What is the magnitude of the torque on the conducting wire by the magnetic field?

\frac{IaB}{b}

IaB IabB

\frac{IB}{ab}

4 solutions

Ritabrata Mukhopadhyay
Mar 26, 2014

According to Fleming's left hand rule side PQ experiences a normal inward force of IaB and the side SR experiences a normal outward force of IaB. As both the forces are equal and opposite they act to form a couple. and moment of couple is torque. Now, Torque = force * perpendicular distance = IaB * bsin 90 = IabB ( sin 90 = 1).

Torque = BIA Sin90 = BIA
According To Question a And b Is the length and the width of the rectangular coil then A = ab Put Value Of A In Equation : Torque = BIA= BIab = IabB

Vivek Sharma - 7 years, 2 months ago

you are absolutely right

Uttaran Choudhurry - 7 years, 1 month ago

Tejas Rangnekar
Apr 25, 2014

$\boxed{Torque=M×B}$ where $\boxed{M=nIA}$ ,A=area

Shahbaz Khan
Apr 6, 2014

here two torques acts t1 and t2 acts along the sides PQ nd RS whose values are t1=f1b/2sin90 and t2=f2b/2sine90 and force , f=ilBsin@ = iaB now t=t1+ t2 =2(ibBa/2) = i(ab)B

Durga Sadasivuni
Mar 20, 2014

Torque= force r sin @ =f*b

here two torques acts t1 and t2 acts along the sides PQ nd RS whose values are t1=f1 a/2 and t2=f2 a/2 and force , f=i l B = i b B now t=t1+ t2 =2 (i b B a/2) = i(ab)B

Torque from magnetic field

4 solutions

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