In Ampere's law `(oint (vec B).(vec dl))=(mu_0)I`, the current outside the curve is not included on the right hand side. Does it mean that the magnetic field B calculated by using Ampere's law , gives the contribution of only the currents crossing the area bounded by the curve?

Ampere's law provides us an easy way to calculate the magnetic field due to a symmetrical distribution of current. Its mathemfield expression is ointvecB.dl=mu_0I_("in") . The quantity on the left hand side is known as line as integral of magnetic field over a closed Ampere's loop. In the above question, find the magnetic field induction at a point distance r from the axis when rgta. Assume relative permeability of the medium surrounding the conductor to be unity.

Ampere's law provides us an easy way to calculate the magnetic field due to a symmetrical distribution of current. Its mathemfield expression is ointvecB.dl=mu_0I_("in") . The quantity on the left hand side is known as line as integral of magnetic field over a closed Ampere's loop. If the current density in a linear conductor of radius a varies with r according to relation J=kr^2 , where k is a constant and r is the distance of a point from the axis of conductor, find the magnetic field induction at a point distance r from the axis when rlta. Assume relative permeability of the conductor to be unity.

Ampere's law provides us an easy way to calculate the magnetic field due to a symmetrical distribution of current. Its mathemfield expression is ointvecB.dl=mu_0I_("in") . The quantity on the left hand side is known as line as integral of magnetic field over a closed Ampere's loop. Only the current inside the Amperian loop contributes in

(A) : The magnetic field at the centre of a circular coil carrying current could be calculated using Amperes law. (R) : Biot savart law could be derived from Amperes law.

A current carrying conducting square frame of side I carrying current I is placed in a uniform transverse magnetic field vec B as shown in the figure. Choose the incorrect statement.

A long solenoid having n = 200 turns per metre has a circular cross-section of radius a_(1) = 1 cm . A circular conducting loop of radius a_(2) = 4 cm and resistance R = 5 (Omega) encircles the solenoid such that the centre of circular loop coincides with the midpoint of the axial line of the solenoid and they have the same axis as shown in Fig. A current 't' in the solenoid results in magnetic field along its axis with magnitude B = (mu)ni at points well inside the solenoid on its axis. We can neglect the insignificant field outside the solenoid. This results in a magnetic flux (phi)_(B) through the circular loop. If the current in the winding of solenoid is changed, it will also change the magnetic field B = (mu)_(0)ni and hence also the magnetic flux through the circular loop. Obvisouly, it will result in an induced emf or induced electric field in the circular loop and an induced current will appear in the loop. Let current in the winding of solenoid be reduced at a rate of 75 A //sec . When the current in the solenoid becomes zero so that external magnetic field for the loop stops changing, current in the loop will follow a differenctial equation given by [You may use an approximation that field at all points in the area of loop is the same as at the centre

A wire of cross-sectional area A forms three sides of a square and is free to rotate about axis OO'. If the structure is deflected by an angle theta from the vertical when current i is passed through it in a magnetic field B acting vertically upward and density of the wire is rho , then the value of theta is given by

A wire of cross-sectional area A forms three sides of a square and is free to rotate about axis OO'. If the structure is deflected by an angle theta from the vertical when current i is passed through it in a magnetic field B acting vertically upward and density of the wire is rho , then the value of theta is given by

A long solenoid having n = 200 turns per metre has a circular cross-section of radius a_(1) = 1 cm . A circular conducting loop of radius a_(2) = 4 cm and resistance R = 5 (Omega) encircles the solenoid such that the centre of circular loop coincides with the midpoint of the axial line of the solenoid and they have the same axis as shown in Fig. A current 't' in the solenoid results in magnetic field along its axis with magnitude B = (mu)ni at points well inside the solenoid on its axis. We can neglect the insignificant field outside the solenoid. This results in a magnetic flux (phi)_(B) through the circular loop. If the current in the winding of solenoid is changed, it will also change the magnetic field B = (mu)_(0)ni and hence also the magnetic flux through the circular loop. Obvisouly, it will result in an induced emf or induced electric field in the circular loop and an induced current will appear in the loop. Let current in the winding of solenoid be reduced at a rate of 75 A //sec . Magentic flux through the loop due to external magnetic field will be I is the current in the loop a_(1) is the radius of solendoid and a_(1)=1cm (given) a_(2) is the radius of circular loop and a_(2)=4cm (given) i is hte current in the solenoid

The basic principle underlying the Hall effect is the Lorentz force. When an electron moves along a direction perpendicular to an applied magnetic field, it experiences a force acting normal to both the directions and moves in response to this force and the force exerted by the internal electric field. For an n -type, "bar" -shaped semiconductor, the carriers are predominantly electrons of bulk density n . We assume that the constant current I flows along the x -axis from left to right in the presence of a magnetic field toward y -axis. Electrons subjected to the lorentz force initially drift away from the current line toward the negative z -axis, resulting in an excess surface electrical charge on the sides of the sample. This charge results in the hall voltage, a potential drop across the two sides of the sample. The transverse voltage is the hall voltage V_(H) and its magnitude is equal to IB//qnd , where I is the current, B is the magnetic field, d is the sample thickness and q is the elementary charge. A silver ribbon lies as shown in the adjacent figure. ( z_(1)=11.8 mm and y_(1)=0.23 mm ) carrying a current of 120 A in the x -direction in a uniform magnetic field B=0.95 T . If electron density is 5.85xx10^(28)//m^(3) , then Magnitude of the drift velocity of elelctrons is