The space within a current carrying solenoid is filled with magneisum having magnetic suscepitbiliy. `x=M_(g) =1.2xx10^(-5)`. What will be the percentage increase in magnetic field?
Text Solution
Verified by Experts
Magnetic field without magneisum is `B_(0)=mu_(0)H` With magnisum, `B=muH=mu_(0)(1+chi)H` `therefore % "increase" =((B)/(B_(0)-1))xx100` `" " =chi_(mg)xx100=1.2xx10^(-5)xx100=1.2xx10^(-3)`
Topper's Solved these Questions
MAGNETISM AND MATTER
DC PANDEY|Exercise 5.1|14 Videos
MAGNETISM AND MATTER
DC PANDEY|Exercise 5.2|14 Videos
MAGNETICS
DC PANDEY|Exercise MCQ_TYPE|1 Videos
MODERN PHYSICS
DC PANDEY|Exercise Integer Type Questions|17 Videos
Similar Questions
Explore conceptually related problems
The space within a current carrying toroid is filled with aluminium of magnetic suceptibiliuty 2.1 xx10^(-5) what is the percentage increase in the magnetic induction B ?
The space within a current carrying toroid is filled with tungsten of susceptibility 4.6 xx 10^(-5) . The percentage increase in the magnetic field is
The space inside a toroid is filled with tungsten whose susceptibility is 6.8xx10^(-5) . The percentage increase in the magnetic field will be
Magnetic Field Due To Current Carrying Solenoid
The percentage increase in magnetic field B when the space within a current carrying toroid is filled with aluminum (chi=2.1xx10^(-5)) is
The magnetic field lines in the middle of the current-carrying solenoid are :