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From a cylinder of radius R, a cylider o...

From a cylinder of radius R, a cylider of radius `R//2` is removed, as shown in Fig. Current flowing in the remaining cylinder is I. Then, magnetic field strength is

A

zero at point A

B

zero at point B

C

`(mu_(0)I)/(3piR)` at point A

D

`(mu_(0) I)/(3 pi R)` at point B

Text Solution

Verified by Experts

The correct Answer is:
C, D
For cylinder ,
`B = (mu_(0) ir)/(2 pi R^(2)) , r lt a`
`= (mu_(0) i)/(2 pi r) ' r ge a`
We can consider the given cyliner as a combination of two cylinders. One of radius 'R' carrying current `I` in one direction and other of radius `(R)/(2)` carrying current `(I)/(3)` in both directions
At point A :
`B=(mu_(0)(I//3))/(2 pi(R//2))+0=(mu_(0)I)/(3 pi R)`
Also point B :
`B = (mu_(0))/(2) ((4I//3)/(pi R^(2)))((R)/(2))+0=(mu_(0)I)/(3 pi R)`.
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Knowledge Check

  • From a cylinder of radius R, a cyclinder of radius R/2 is removed, as shown . Current flowing in the remaning cylinder is l. magnetic field strength is :

    A
    zero at point A
    B
    zero at point B
    C
    `(mu_(0)I)/(3piR)` at point A
    D
    `(mu_(0)I)/(3piR)` at point B

  • A long wire bent as shown in fig. carries current I. If the radius of the semicircular portion is a, the magnetic field at centre C is

    A
    `(mu_0I)/(4a)`
    B
    `(mu_0I)/(4pia)sqrt(pi^2+4)`
    C
    `(mu_0I)/(4a)+(mu_0I)/(4pia)`
    D
    `(mu_0I)/(4pia)sqrt((pi^2-4))`

  • An infinity long cylinder of radius R has an infinitely long cylindrical cavity of radius (R)/(2) are shown in the figure. The remaining portion has uniform volume charge density rho . The magnitude of electrical field at the centre of the cavity is-

    A
    `(rhoR)/(6epsilon_(0))`
    B
    `(rhoR)/(4epsilon_(0))`
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