Integer Answer Type Questions: Moving Charges & Magnetism | JEE Advanced | 35 Years Chapter wise Previous Year Solved Papers for JEE PDF Download

Q.1. A steady current I goes through a wire loop PQR having shape of a right angle triangle with PQ = 3x, PR = 4x and QR = 5x. If the magnitude of the magnetic field at P due to this loop is  find the value of k.

Ans. 7

Solution. 

The right angled triangle is shown in the figure. Let us drop a perpendicular from P on QR which cuts QR at M.

The magnatic field due to currents in PQ and RP at P is zero.

The magnetic field due to current in QR at P is

In ΔPRM,

16x² = PM² + (5x – a)² … (iii)

⇒ 7 x² = 25 x²- 10xa ⇒  10xa =  18x²

⇒  a = 1.8 x … (iv)

From (ii) & (iv),

9 x² = PM²+ (1.8x)²

From (i), (v), (vi) and (vii),

Comparing it with B =

Q.2. A long circular tube of length 10 m and radius 0.3 m carries a current I along its curved surface as shown. A wire-loop of resistance 0.005 ohm and of radius 0.1 m is placed inside the tube with its axis coinciding with the axis  of the tube. The current varies as I = I₀cos(300 t) where I₀ is constant. If the magnetic moment of the loop is Nμ₀I₀sin (300 t), then ‘N’ is

Ans. 6

Solution. Let us consider an amperian loop ABCD which is a rectangle as shown in the figure. Applying ampere’s circuital law we get

 (current passing through the loop)

The magnetic moment of the loop = (current in the loop) × πr²

Comparing it with the expression given in the question we get

Q.3.A cylindrical cavity of diameter a exists inside a cylinder of diameter 2a as shown in the figure. Both the cylinder and the cavity are infinity long. A uniform current density J flows along the length. If the magnitude of the magnetic field at the point P is given by  then the value of N is

Ans.5

Solution. Current density J =

Let us consider the cavity to have current I' flowing in both the directions.

The magnetic field at P due to the current flowing through the cylinder

The magnetic field at P due to the current (I’) flowing in opposite direction is

∴ The net magnetic field is

Q.4. Two parallel wires in the plane of the paper are distance X₀ apart. A point charge is moving with speed u between the wires in the same plane at a distance X₁ from one of the wires. When the wires carry current of magnitude I in the same direction, the radius of curvature of the path of the point charge is R₁. In contrast, if the currents I in the two wires have directions opposite to each other, the radius of curvature of the path is R₂. If   the value of   is

Ans. 3

Solution.