A small coil C with $$N=$$ $$200$$ turns is mounted on one end of a balance beam and introduced between the poles of an electromagnet as shown in figure. The area of the coil is S $$=$$ $$1cm2$$, the length of the right arm of the balance beam is I $$=$$ $$30$$ cm. When there is no current in the coil the balance is in equilibrium. On passing a current I $$=$$ $$22$$ mA through the coil, equilibrium is restored by putting an additional weight of mass m $$=$$ $$60$$ mg on the balance pan. Find the magnetic induction field (in$$ terms of x $$10-1$$ $$T) between the poles of the electromagnet, assuming it to be uniform.

Question

Infinity Learn · Accepted Answer

Magnetic torque $$=NISBsin(S$$→,B→$$)$$  NISB $$sin90$$°  Gravitational torque $$=($$Δm×$$g)l$$  For equilibrium $$NISB=$$Δmgl⇒ $$B=$$Δ$$mgl/NIS$$ $$B=60$$×$$10-6$$×$$9.8$$×$$0.3200$$×$$22$$×$$10-3$$×$$1$$×$$10-4=4$$×$$10-1$$ T

Magnetic field due to current element

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$Magnetic torque = NISB \sin (\vec{S}, \vec{B}) NISB \sin 90 ° Gravitational torque = (Δ m \times g) l For equilibrium NISB = Δ mgl \Rightarrow B = Δ m g l / N I S B = \frac{(60 \times 10^{- 6}) \times 9.8 \times 0.3}{200 \times 22 \times 10^{- 3} \times 1 \times 10^{- 4}} = 4 \times 10^{- 1} T$

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Magnetic field due to current element

Remember concepts with our Masterclasses.

Magnetic torque =NISBsin(S→,B→) NISB sin90° Gravitational torque =(Δm×g)l For equilibrium NISB=Δmgl⇒ B=Δmgl/NIS B=60×10-6×9.8×0.3200×22×10-3×1×10-4=4×10-1 T

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