13 - Moving charge magnetism and force in magnetic field Questions Answers

Please post solution sir, thanks in advance. 
Asked By: TUSHAR GUPTA
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Thanks in advance for solution. 
Asked By: TUSHAR GUPTA
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Joshi sir comment
Thanks in advance 
Asked By: TUSHAR GUPTA
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Asked By: TUSHAR GUPTA
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Asked By: KESHAV
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Asked By: LUFFY
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Joshi sir comment

Magnetic torque = M×B=iA×B Torque due to gravity about the point of  circumference = R×mg Now compare the two   Inform for any issue

CALCULATE  THE MAGNETIC FIELD  AT THE CENTRE OF SQUARE WIRE OF SIDE (2)1/2cm  AND CARRYING A CURRENT OF 10A.?

Asked By: GAUTAM SAGAR
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Joshi sir comment

method given by BONEY is correct but formula is not correct

it should be B = 4×{ (μ0I) / 4πR } × [sinθ1+ Sinθ2]

two short magnets of equal dipole momet M are fastened perpedicularat their centre.the magnitude of  magnetic feild at adistance d from the centre on the bisector of right angles is?

Asked By: AKANKSHA Solved By: SARIKA
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Joshi sir comment

the resultant dipole moment is √2M  thus  the resultant magnetic  field is  μ2√2M/4πd3   (ans)

this answer is given by sarika and is correct answer

 

velocity of e- in hydrogen atom is 2X106 m/s . the radius of orbit is 5X10-11 m . the mag induction at centre of orbit in T will be??

Asked By: SARIKA 1 Month ago
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Asked By: SARIKA SHARMA
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Joshi sir comment

B = μ0/2 * i/r

and i = e/T

and T = 2πr/v

solve

potential diff b/w dees of cyclotron is V , if charge q comes out of it after completing n revolution then gain in KE of charge is 2nqV HOW?

Asked By: SARIKA 1 Month ago
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Asked By: SARIKA SHARMA
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Joshi sir comment

when charge will move from first dee to second dee it will gain qV and on returning to first dee it will gain qV again so in one revolution it will gain 2qV so in n revolutions it will gain 2qVn

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