Question 1:
easyIf a long hollow copper pipe carries a direct current, the magnetic field associated with the current will be :
Inside the pipe current enclosed will be zero so magnetic field will also be zero. Outside the pipe there will be net Magnetic field
Question 2:
easyIn a co-axial straight cable, the central conductor and the outer conductor have equal currents in opposite directions. The magnetic induction is zero :
As outside the conductor net current enclosed is zero. using ampere circuital law Magnetic field is also zero.
Question 3:
easyA solenoid consists of 100 turns of wire and has a length of 10.0cm. The magnetic field inside the solenoid when it carries a current of 0.500 A will be :
\[Â B= \mu_{0}ni \] where n is number of turns per unit length
n= 100/0.1=1000
i=0.5
substituting we get \[ B = 6.28\times 10^{-4} T\]
Question 4:
easyFor the hollow thin cylindrical current carrying straight pipe which statement is correct:
Inside the hollow pipe, the magnetic field is zero according to Ampere's law. Since a current-carrying pipe is electrically neutral, the electric field outside the pipe is zero.
Question 5:
easyAssertion (A): A rectangular current loop is in an arbitrary orientation in an external uniform magnetic field. No work is required to rotate the loop about an axis perpendicular to its plane.
Reason (R): All positions represent the same level of energy.
Assertion (A): A current loop in a uniform magnetic field experiences a torque \(\vec{\tau} = \vec{M} \times \vec{B}\). Work is generally required to change its orientation. So, (A) is false. Reason (R): The potential energy of a current loop in a magnetic field is \(U = -\vec{M} \cdot \vec{B}\), which depends on the orientation of \(\vec{M}\) relative to \(\vec{B}\). Thus, not all positions represent the same energy. So, (R) is false. Both (A) and (R) are false.
Question 6:
easyAssertion (A): In Ampere’s law for magnetostatics \(\oint \vec{B} \cdot d\vec{l} = \mu_0 \sum I_{\text{i}}\) the current outside the Amperian loop is not included on the right side.
Reason (R): Magnetic field calculated using Ampere’s law is due to inside as well outside the current of closed loop.
Assertion (A): Ampere's law \(\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enclosed}}\) states that only currents passing through the Amperian loop contribute to the right-hand side. So, (A) is true.
Reason (R): The magnetic field (vec{B}) on the left-hand side of Ampere's law is the total field produced by all currents, both inside and outside the loop. So, (R) is true. However, R describes the nature of (vec{B}), not why only enclosed currents are counted on the right side. Thus, (R) is not the correct explanation of (A).
Question 7:
easyA long solenoid of radius $1\text{ mm}$ has $100$ turns per mm. If $1\text{ A}$ current flows in the solenoid, the magnetic field strength at the centre of the solenoid is:
(2022)
The magnetic field inside a long solenoid is given by $B = \mu_0 n I$. Given $n = 100\text{ turns/mm} = 10^5\text{ turns/m}$ and $I = 1\text{ A}$, substituting the values yields $B = (4\pi \times 10^{-7}) \times 10^5 \times 1 = 12.56 \times 10^{-2}\text{ T}$.
Question 8:
easyA long solenoid carrying a current produces a magnetic field $B$ along its axis. If the current is doubled and the number of turns per cm is halved, the new value of the magnetic field is:
(2003)
The magnetic field of a solenoid is $B = mu_0 n I$. When the current is doubled ($I' = 2I$) and turns per unit length are halved ($n' = n/2$), the new magnetic field is $B' = mu_0 (n/2)(2I) = mu_0 n I = B$.