1.

A uniform magnetic needle of strength of each pole is 98.1 amp, cm, is suspended from its centre by a thread. When a mass of 50 mg is loaded to its upper end, the needle becomes horizontal, then the vertical component of earth's magnetic induction is (g=981 cm//s^(2))

Answer»

0.50 gauss
0.25 gauss
0.05 gauss
0.005 gauss

Solution :
Taking MOMENT about O,
`Mgl=mB_(V)L+mB_(v)L`
`Mg=2mB_(v)`
`(50)/(1000)xx981 = 2xx98.1xxB_(v)`
`B_(v)=0.25` gauss


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