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Find the experssion for the magnetic field at the centre O of a coil bent in the form of a square of side `2a`, carrying current I, figure. |
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Answer» Here, `OE=AE=a`, `/_AOE=/_BOE=45^@` Magnetic field at O due to current in arm AB of wire `B_1=(mu_0)/(4pi)I/a[sin 45^@+sin45^@]=(sqrt2mu_0I)/(4pia)` It is acting vertically downwards. The magnetic field at O due to currents in arms BC, CD and DA will also be of the same magnitude and direction as `vecB_1`. Therefore, resultant magnetic field at O is `B=4B_1=(4xxsqrt2mu_0I)/(4pia)=(sqrtmu_0I)/(pia)` It is acting vertically downwards. |
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