# 26 - Magnetism Questions Answers

**Asked By: ROHITGARG**6 year ago

**Solved By: ARICA**

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A bar magnet of magnetic moment '**m' **and moment of inerta** 'I' **(about an axis passing through centre of mass and perpendicular to the magnet) is turned through 90^{0} about the axis of suspension, to be at right angle to a uniform magnetic field B and then released. What is the __angular velocity of the magnet__while passing through the direction of the field ??

**Asked By: VIRAT**6 year ago

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A SIMPLE ELECTRIC MOTOR HAS AN ARMATURE RESISTANCE OF 1 ohm & RUNS FROM A D.C SOURCE OF 12V . IT DRAWS A CURRENT OF 2A WHEN UNLOADED . WHEN A CERTAIN LOAD IS CONNECTED TO IT ,IT SPEED REDUCES BY 10% OF ITS INITIAL VALUE ,THE CURRENT DRAWN BY THE UNLOADED MOTOR IS??

ANS 3A

**Asked By: SARIKA SHARMA**15 Day ago

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SIR , ISTHE FOLLOWG METH. IS CORRECT

R=1ohm , E=12V , I=2A

I=(E-e)/R => e=10 ON PUTTING D VALUES

NOW AS IT IS REDUCES BY 10% MEANS NOW e' = 9V ( 10/100 X 10 = 1 CHANGE , THUS FINAL = 10-1=9)

I= E-e'/R = ( 12-9)/ 1 = 3A (RESISTANCE IS KEPT CONST.) RESISTANCE WILL CHANGE OR NOT??

**SARIKA SHARMA**15 Day ago

**Asked By: SARIKA**7 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

yes this method is correct.

is megnatic & electric feild depend on the frame give proper illustration please?

**Asked By: RISHABH GOUR**7 year ago

**Solved By: SARIKA**

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Q A SIMPLE ELECTRIC MOTOR HAS AN ARMATURE RESISTANCE OF 1 ohm & RUNS FROM A D.C SOURCE OF 12V . IT DRAWS A CURRENT OF 2A WHEN UNLOADED . WHEN A CERTAIN LOAD IS CONNECTED TO IT ,IT SPEED REDUCES BY 10% OF ITS INITIAL VALUE ,THE CURRENT DRAWN BY THE UNLOADED MOTOR IS??

ANS 3A

**Asked By: SARIKA SHARMA**7 year ago

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velocity of e^{-} in hydrogen atom is 2X10^{6} 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

**Asked By: SARIKA SHARMA**7 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

B = μ_{0}/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**7 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

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

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**7 year ago

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velocity of e^{-} in hydrogen atom is 2X10^{6} m/s . the radius of orbit is 5X10^{-11} m . the mag induction at centre of orbit in T will be??

**Asked By: SARIKA**7 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

B = μ_{0}/2 * i/r

i = e/T

and T = 2πr/v

now solve