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Chapter 26: Problem 54

A single-turn wire loop \(10 \mathrm{cm}\) in diameter carries a 12 - A current. It experiences a \(0.015 \mathrm{N} \cdot \mathrm{m}\) torque when the normal to the loop plane makes a \(25^{\circ}\) angle with a uniform magnetic field. Find the magnetic field strength.

Short Answer

Expert verified
The strength of the magnetic field is approximately 2.8 T.

Step by step solution

01

Identify the known quantities

From the problem, we know that the torque (\(\tau\)) is 0.015 Nm, the current (I) is 12 A, the radius of the loop (r) is 0.05 m (since the diameter is 10 cm, the radius would be half of it), and the angle (\(\theta\)) is \(25^{\circ}\). We can convert the angle from degrees to radians for easier manipulation.
02

Find the loop's area (A)

The area of a circle is given by \(\pi r^2\), substituting the known radius we get \(A = \pi (0.05)^2\) m².
03

Convert the given angle to radians

\(\theta = 25^{\circ} = 25*(\pi/180)\) rad, or approximately 0.435 rad.
04

Rearrange the formula and substitute the known quantities

The formula for torque can be rearranged to solve for B the magnetic field: \(B = \tau / (IA \sin(\theta))\). Substituting the known quantities we get \(B = 0.015 Nm / ((12 A)*(\pi (0.05)^2 m²)*(\sin(0.435 rad)))\).
05

Solve for B

Doing the necessary computations, we find that \(B \approx 2.8 T\).

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